🧠:<think>Okay, so I have these two problems to solve. Let me take them one at a time. Starting with the first problem: The author allocates time for Fiction and Non-Fiction in the ratio 3:2 respectively, and the total time is 25 hours per week. I need to find out how much time is allocated to each genre.Hmm, ratios. I remember that ratios can be converted into fractions of the total. The ratio is 3:2, which means for every 3 parts of Fiction, there are 2 parts of Non-Fiction. So, the total number of parts is 3 + 2 = 5 parts. If 5 parts equal 25 hours, then each part must be 25 divided by 5. Let me calculate that: 25 ÷ 5 = 5. So each part is 5 hours. Therefore, Fiction, which is 3 parts, would be 3 × 5 = 15 hours. And Non-Fiction, being 2 parts, would be 2 × 5 = 10 hours. Wait, let me double-check. 15 + 10 is 25, which matches the total given. So that seems correct. Alright, moving on to the second problem. The author's success is modeled by a quadratic function: S(t) = -2t² + 12t + 50. I need to find the maximum success S and the corresponding time t.Quadratic functions. I remember that a quadratic function in the form of S(t) = at² + bt + c has its vertex at t = -b/(2a). Since the coefficient of t² is negative (-2), the parabola opens downward, meaning the vertex is the maximum point. So, that's where the maximum success occurs.Let me write down the formula for the vertex. t = -b/(2a). In this case, a is -2 and b is 12. Plugging those in: t = -12/(2*(-2)) = -12/(-4) = 3. So, t is 3 hours. Now, to find the maximum success S, I need to plug t = 3 back into the original equation. Calculating S(3): S = -2*(3)² + 12*(3) + 50. Let's compute each term step by step.First, (3)² is 9. Multiply that by -2: -2*9 = -18.Next, 12*3 is 36.So now, S = -18 + 36 + 50. Adding those together: (-18 + 36) is 18, and 18 + 50 is 68.Therefore, the maximum success is 68 when t is 3 hours.Wait, let me verify my calculations. Compute S(3) again:-2*(3)^2 = -2*9 = -1812*3 = 36So, -18 + 36 = 1818 + 50 = 68. Yep, that's correct.Alternatively, I can use calculus to find the maximum. Taking the derivative of S(t) with respect to t:S'(t) = d/dt (-2t² + 12t + 50) = -4t + 12.Setting the derivative equal to zero for critical points:-4t + 12 = 0-4t = -12t = 3. So, same result. Then, plugging t=3 into S(t) gives 68. So, that's consistent.Alternatively, completing the square might also work, but since I already have the vertex formula, it's quicker.Just to recap: For a quadratic function, the vertex is at t = -b/(2a). Plugging in the values gives t=3, and substituting back gives S=68. So, that's the maximum success.I think that's all for both problems. The first one was straightforward with ratios, and the second one required using the vertex formula for a quadratic function. Both seem to check out.Final Answer1. Fiction: boxed{15} hours, Non-Fiction: boxed{10} hours.2. Maximum success is boxed{68} achieved at boxed{3} hours.</think>

🧠:<think>Okay, so I need to figure out the total projected rental income for the next year for all 10 apartments. The apartments are split between Midtown and Heights. There are 6 in Midtown and 4 in Heights. I have data on average monthly rent, annual appreciation rates, occupancy rates, and maintenance costs. Also, there's a second part where I might sell one Midtown apartment and buy one in Heights, which would affect the total income. Let me break this down step by step.First, for part 1, I need to calculate the projected rental income for each neighborhood separately and then combine them. I should consider the appreciation, occupancy, and subtract maintenance costs.Starting with Midtown. There are 6 apartments. The average monthly rent is 1,500. But there's an annual appreciation rate of 4%. So, the rent next year will be higher. I think I need to calculate the new rent after appreciation. Appreciation is usually on the property value, but here it's mentioned as the average monthly rent, so I assume it's the rent that's appreciating. So, the new rent will be 1,500 multiplied by 1.04. Let me compute that: 1,500 * 1.04 = 1,560. So, next year's rent per month in Midtown will be 1,560.But wait, is the appreciation applied once annually, so for the next year, it's just a 4% increase? Yes, that's correct. So, the rent goes up by 4% each year.Now, the occupancy rate in Midtown is 92%. That means, on average, 92% of the apartments are rented out each month. So, for 6 apartments, the average number of occupied units is 6 * 0.92 = 5.52. Since you can't have a fraction of an apartment, but for calculation purposes, we can keep it as 5.52.So, the monthly rental income from Midtown would be 5.52 apartments * 1,560 per month. Let me calculate that: 5.52 * 1,560. Hmm, 5 * 1,560 is 7,800, and 0.52 * 1,560 is 811.2, so total is 7,800 + 811.2 = 8,611.2 per month. Then, for the year, that's 8,611.2 * 12. Let me compute that: 8,611.2 * 12. 8,000 * 12 is 96,000, 611.2 * 12 is 7,334.4, so total is 96,000 + 7,334.4 = 103,334.4 dollars per year.But wait, that's just the rental income. I need to subtract the maintenance costs. Maintenance is 200 per apartment per month. So, for Midtown, 6 apartments, that's 6 * 200 = 1,200 per month. For the year, that's 1,200 * 12 = 14,400.So, Midtown's net income is 103,334.4 - 14,400 = 88,934.4 dollars.Now, moving on to Heights. There are 4 apartments. The average monthly rent is 1,300 with a 3% appreciation. So, next year's rent will be 1,300 * 1.03 = 1,339 per month.Occupancy rate is 88%, so the average number of occupied units is 4 * 0.88 = 3.52.Monthly rental income is 3.52 * 1,339. Let me compute that: 3 * 1,339 is 4,017, and 0.52 * 1,339 is approximately 700.28. So, total is 4,017 + 700.28 = 4,717.28 per month. For the year, that's 4,717.28 * 12. Let me calculate: 4,700 * 12 is 56,400, and 17.28 * 12 is 207.36, so total is 56,400 + 207.36 = 56,607.36 dollars.Subtracting maintenance costs: 4 apartments * 200 per month = 800 per month. For the year, that's 800 * 12 = 9,600.So, Heights' net income is 56,607.36 - 9,600 = 47,007.36 dollars.Now, total projected rental income for all 10 apartments is Midtown net + Heights net: 88,934.4 + 47,007.36 = 135,941.76 dollars.Wait, but the question says "projected rental income for the next year," so I think that's correct. But let me double-check my calculations.Midtown: 6 apartments, 92% occupancy, 4% appreciation.Rent next year: 1,500 * 1.04 = 1,560.Occupied units: 6 * 0.92 = 5.52.Monthly income: 5.52 * 1,560 = 8,611.2.Yearly: 8,611.2 * 12 = 103,334.4.Maintenance: 6 * 200 * 12 = 14,400.Net: 103,334.4 - 14,400 = 88,934.4.Heights: 4 apartments, 88% occupancy, 3% appreciation.Rent next year: 1,300 * 1.03 = 1,339.Occupied units: 4 * 0.88 = 3.52.Monthly income: 3.52 * 1,339 ≈ 4,717.28.Yearly: 4,717.28 * 12 ≈ 56,607.36.Maintenance: 4 * 200 * 12 = 9,600.Net: 56,607.36 - 9,600 = 47,007.36.Total net: 88,934.4 + 47,007.36 = 135,941.76.So, approximately 135,941.76.But the question mentions "projected rental income," which I think includes the appreciation, so this should be correct.Now, moving on to part 2. If I sell one apartment in Midtown and buy one in Heights, how does this affect the total projected rental income? There's a one-time cost of 5,000 for selling and buying, which should be deducted from the total rental income.So, first, let's adjust the number of apartments: Midtown goes from 6 to 5, and Heights goes from 4 to 5.I need to recalculate the net income with 5 in Midtown and 5 in Heights, then subtract 5,000.Let me compute Midtown first with 5 apartments.Midtown: 5 apartments, 92% occupancy, 4% appreciation.Rent next year: 1,500 * 1.04 = 1,560.Occupied units: 5 * 0.92 = 4.6.Monthly income: 4.6 * 1,560 = let's compute that: 4 * 1,560 = 6,240, 0.6 * 1,560 = 936, so total 6,240 + 936 = 7,176 per month.Yearly: 7,176 * 12 = 86,112.Maintenance: 5 * 200 * 12 = 12,000.Net: 86,112 - 12,000 = 74,112.Heights: Now 5 apartments, 88% occupancy, 3% appreciation.Rent next year: 1,300 * 1.03 = 1,339.Occupied units: 5 * 0.88 = 4.4.Monthly income: 4.4 * 1,339 ≈ let's calculate: 4 * 1,339 = 5,356, 0.4 * 1,339 = 535.6, so total 5,356 + 535.6 = 5,891.6 per month.Yearly: 5,891.6 * 12 ≈ 70,700 (exactly: 5,891.6 * 12 = 70,700 - wait, 5,891.6 * 10 = 58,916, plus 5,891.6 * 2 = 11,783.2, so total 58,916 + 11,783.2 = 70,699.2).Maintenance: 5 * 200 * 12 = 12,000.Net: 70,699.2 - 12,000 = 58,699.2.Total net income after the change: 74,112 (Midtown) + 58,699.2 (Heights) = 132,811.2.But we have to subtract the one-time cost of 5,000 for selling and buying. So, total projected rental income becomes 132,811.2 - 5,000 = 127,811.2.Comparing to the original total of 135,941.76, the change is a decrease of 135,941.76 - 127,811.2 = 8,130.56.So, the total projected rental income decreases by approximately 8,130.56.Wait, let me verify the calculations again.Midtown with 5 apartments:Rent: 1,560.Occupied: 4.6.Monthly income: 4.6 * 1,560 = 7,176.Yearly: 7,176 * 12 = 86,112.Maintenance: 5 * 200 * 12 = 12,000.Net: 86,112 - 12,000 = 74,112.Heights with 5 apartments:Rent: 1,339.Occupied: 4.4.Monthly income: 4.4 * 1,339 = 5,891.6.Yearly: 5,891.6 * 12 = 70,699.2.Maintenance: 5 * 200 * 12 = 12,000.Net: 70,699.2 - 12,000 = 58,699.2.Total net: 74,112 + 58,699.2 = 132,811.2.Subtract 5,000: 132,811.2 - 5,000 = 127,811.2.Original total was 135,941.76, so the difference is 135,941.76 - 127,811.2 = 8,130.56.So, the total projected rental income decreases by approximately 8,130.56.Alternatively, we can express this as a decrease of about 8,131.But let me check if I considered the appreciation correctly. The appreciation is annual, so for the next year, the rent increases by 4% and 3% respectively. So, yes, that's correct.Also, the occupancy rates are applied to the number of apartments, which is correct.Maintenance is per apartment per month, so that's correctly calculated as 200 * number of apartments * 12.So, the calculations seem correct.Therefore, the answers are:1. Total projected rental income: approximately 135,941.76.2. After selling one Midtown and buying one Heights, total projected rental income becomes approximately 127,811.2, which is a decrease of about 8,130.56.But the question asks for the total projected rental income after the change, so it's 127,811.2, minus the 5,000 cost, which we already did.Wait, no, the 5,000 is a one-time cost, so it's subtracted from the total rental income. So, the total rental income is 132,811.2 - 5,000 = 127,811.2.Yes, that's correct.So, summarizing:1. Original total: 135,941.76.2. After rebalancing: 127,811.2.The difference is a decrease of 8,130.56.I think that's it.Final Answer1. The total projected rental income for the next year is boxed{135941.76} dollars.2. After rebalancing, the total projected rental income is boxed{127811.20} dollars.</think>

🧠:<think>Okay, so I have two questions here related to video game sales and nostalgia factors. Let me try to tackle them one by one.Starting with the first question: It's about calculating the average sales of a new Pokémon-inspired game over the first 12 years. The sales function is given as ( S(t) = 31 times left(1 + frac{1}{20} sinleft(frac{pi}{6}tright)right) ). I need to find the average sales over the first 12 years. Hmm, average value of a function over an interval. I remember that the average value of a function ( f(t) ) over the interval [a, b] is given by ( frac{1}{b - a} int_{a}^{b} f(t) dt ). So in this case, a is 0 and b is 12. Therefore, the average sales ( overline{S} ) would be ( frac{1}{12 - 0} int_{0}^{12} S(t) dt ).Let me write that down:( overline{S} = frac{1}{12} int_{0}^{12} 31 left(1 + frac{1}{20} sinleft(frac{pi}{6}tright)right) dt )I can factor out the 31:( overline{S} = frac{31}{12} int_{0}^{12} left(1 + frac{1}{20} sinleft(frac{pi}{6}tright)right) dt )Now, let's split the integral into two parts:( overline{S} = frac{31}{12} left( int_{0}^{12} 1 dt + frac{1}{20} int_{0}^{12} sinleft(frac{pi}{6}tright) dt right) )Calculating the first integral is straightforward. The integral of 1 with respect to t from 0 to 12 is just 12.For the second integral, I need to integrate ( sinleft(frac{pi}{6}tright) ). The integral of ( sin(k t) ) is ( -frac{1}{k} cos(k t) ). So here, k is ( frac{pi}{6} ), so the integral becomes ( -frac{6}{pi} cosleft(frac{pi}{6}tright) ).Let me compute that:First integral: ( int_{0}^{12} 1 dt = 12 )Second integral: ( frac{1}{20} times left[ -frac{6}{pi} cosleft(frac{pi}{6}tright) right]_0^{12} )Let me compute the bounds:At t = 12: ( cosleft(frac{pi}{6} times 12right) = cos(2pi) = 1 )At t = 0: ( cos(0) = 1 )So plugging in:( frac{1}{20} times left( -frac{6}{pi} [1 - 1] right) = frac{1}{20} times 0 = 0 )Wait, that's interesting. So the second integral evaluates to zero.Therefore, the average sales become:( overline{S} = frac{31}{12} times (12 + 0) = frac{31}{12} times 12 = 31 )So the average sales over the first 12 years is 31 million copies. That's the same as the original games. Hmm, that makes sense because the sine function oscillates around zero, so its average over a full period is zero. Therefore, the average sales just equal the base sales, which is 31 million.Okay, that seems straightforward. Let me just verify my steps:1. Set up the average value formula correctly.2. Split the integral into two parts.3. Calculated the first integral as 12.4. For the second integral, recognized the sine function's integral, applied the limits, and saw that the cosine terms at 0 and 12 are both 1, so their difference is zero.5. Thus, the second integral contributes nothing to the average, leaving just the base sales.Yes, that seems correct.Moving on to the second question: It's about determining the age at which the nostalgia factor drops below 20. The function given is ( N(a) = 100e^{-0.05a} ), where N(a) is the nostalgia factor on a scale from 0 to 100, and a is the age in years. We need to find the age a when N(a) < 20.So, set up the inequality:( 100e^{-0.05a} < 20 )Divide both sides by 100:( e^{-0.05a} < 0.2 )Take the natural logarithm of both sides. Remember that ln is a monotonically increasing function, so the inequality direction remains the same when taking logs.( ln(e^{-0.05a}) < ln(0.2) )Simplify the left side:( -0.05a < ln(0.2) )Now, solve for a. Since we're dividing by a negative number (-0.05), the inequality sign will reverse.( a > frac{ln(0.2)}{-0.05} )Compute the right side:First, calculate ( ln(0.2) ). I know that ln(0.2) is negative because 0.2 is less than 1. Let me compute it:( ln(0.2) approx -1.6094 )So,( a > frac{-1.6094}{-0.05} = frac{1.6094}{0.05} )Divide 1.6094 by 0.05:1.6094 / 0.05 = 32.188So, a > approximately 32.188 years.Since age is a continuous variable, the nostalgia factor drops below 20 at approximately 32.19 years. Depending on how precise we need to be, we can round it to two decimal places or more. But since the question asks for the age, I think two decimal places are sufficient, so 32.19 years.Let me verify my steps:1. Set up the inequality correctly: 100e^{-0.05a} < 202. Divided both sides by 100: e^{-0.05a} < 0.23. Took natural logs: -0.05a < ln(0.2)4. Remembered to reverse the inequality when dividing by negative: a > ln(0.2)/(-0.05)5. Calculated ln(0.2) ≈ -1.60946. Divided by -0.05: 32.1887. Rounded to two decimal places: 32.19Yes, that seems correct.But just to be thorough, let me plug a = 32.19 back into N(a):N(32.19) = 100e^{-0.05 * 32.19} ≈ 100e^{-1.6095} ≈ 100 * 0.2 ≈ 20So at a ≈ 32.19, N(a) is approximately 20. Therefore, for ages above 32.19, N(a) drops below 20.Alternatively, if we use more precise calculations, ln(0.2) is approximately -1.60943791, so:a = (-1.60943791)/(-0.05) = 32.1887582So, approximately 32.19 years. So, the age is about 32.19 years.Therefore, the nostalgia factor drops below 20 at approximately 32.19 years of age.I think that's solid.Final Answer1. The average sales over the first 12 years is boxed{31} million copies.2. The nostalgia factor drops below 20 at an age of approximately boxed{32.19} years.</think>

🧠:<think>Okay, so I have this problem where an investor is looking at a new product line. They've given me two functions: one for revenue and one for cost. My job is to figure out how many units they need to sell to maximize profit, and then determine the range of units that gives them at least a 30% profit margin. Hmm, let's break this down step by step.First, part 1: finding the number of units x that maximizes the profit P(x). Profit is defined as revenue minus cost, so I need to write out the profit function. The revenue function is R(x) = 500x - 0.5x², and the cost function is C(x) = 200x + 3000. So, profit P(x) should be R(x) - C(x). Let me write that out:P(x) = R(x) - C(x) = (500x - 0.5x²) - (200x + 3000)Simplify that:P(x) = 500x - 0.5x² - 200x - 3000Combine like terms:500x - 200x is 300x, so:P(x) = -0.5x² + 300x - 3000Alright, so now I have a quadratic function for profit. Quadratic functions have the form ax² + bx + c, and since the coefficient of x² is negative (-0.5), this parabola opens downward. That means the vertex of the parabola is the maximum point. So, the x-value at the vertex will give me the number of units that maximizes profit.To find the vertex of a quadratic function, the formula is x = -b/(2a). In this case, a is -0.5 and b is 300. Let me plug those in:x = -300 / (2 * -0.5) = -300 / (-1) = 300So, x = 300 units. That should be the number of units that maximizes profit. But wait, let me make sure I did that correctly. The formula is x = -b/(2a). So, a is -0.5, so 2a is -1, and b is 300, so -b is -300. So, -300 divided by -1 is indeed 300. Okay, that seems right.But just to be thorough, maybe I can check the second derivative to confirm it's a maximum. The first derivative of P(x) is P'(x) = d/dx (-0.5x² + 300x - 3000) = -x + 300. Setting that equal to zero gives -x + 300 = 0, so x = 300. That's the critical point.Then, the second derivative is P''(x) = d/dx (-x + 300) = -1. Since the second derivative is negative (-1), that means the function is concave down at that point, confirming it's a maximum. Okay, so that's solid. 300 units will maximize the profit.Moving on to part 2: the investor wants a minimum profit margin of 30% on the revenue. So, profit margin is profit divided by revenue, right? So, the condition is P(x)/R(x) ≥ 0.30.Let me write that out:P(x)/R(x) ≥ 0.30We already have expressions for P(x) and R(x). Let's substitute them in:(-0.5x² + 300x - 3000)/(500x - 0.5x²) ≥ 0.30Hmm, that's a bit complex. Let me write it as:(-0.5x² + 300x - 3000) ≥ 0.30*(500x - 0.5x²)Let me compute the right-hand side:0.30*(500x - 0.5x²) = 150x - 0.15x²So, now the inequality becomes:-0.5x² + 300x - 3000 ≥ 150x - 0.15x²Let me bring all terms to the left side:-0.5x² + 300x - 3000 - 150x + 0.15x² ≥ 0Combine like terms:(-0.5x² + 0.15x²) + (300x - 150x) + (-3000) ≥ 0Calculating each part:-0.5 + 0.15 is -0.35, so:-0.35x² + 150x - 3000 ≥ 0So, we have:-0.35x² + 150x - 3000 ≥ 0Hmm, let's multiply both sides by -1 to make the coefficient of x² positive. But remember, multiplying an inequality by a negative number reverses the inequality sign.So:0.35x² - 150x + 3000 ≤ 0Now, we have a quadratic inequality: 0.35x² - 150x + 3000 ≤ 0To solve this, first, let's find the roots of the quadratic equation 0.35x² - 150x + 3000 = 0.Using the quadratic formula:x = [150 ± sqrt(150² - 4*0.35*3000)] / (2*0.35)First, compute discriminant D:D = 150² - 4*0.35*3000150² is 22500.4*0.35 is 1.4, and 1.4*3000 is 4200.So, D = 22500 - 4200 = 18300So, sqrt(18300). Let me compute that.18300 is 100*183, so sqrt(18300) = 10*sqrt(183). Let me approximate sqrt(183). 13² is 169, 14² is 196, so sqrt(183) is between 13 and 14. Let's see, 13.5² is 182.25, which is very close to 183. So, sqrt(183) ≈ 13.53.Therefore, sqrt(18300) ≈ 10*13.53 = 135.3So, x = [150 ± 135.3]/(2*0.35) = [150 ± 135.3]/0.7Compute both roots:First root: (150 + 135.3)/0.7 = 285.3/0.7 ≈ 407.57Second root: (150 - 135.3)/0.7 = 14.7/0.7 ≈ 21So, the quadratic equation has roots at approximately x ≈ 21 and x ≈ 407.57.Since the coefficient of x² is positive (0.35), the parabola opens upwards. Therefore, the quadratic expression 0.35x² - 150x + 3000 is ≤ 0 between its roots.So, the solution to the inequality is x between 21 and approximately 407.57.But wait, we need to make sure that x is within the domain where revenue is positive, right? Because revenue R(x) = 500x - 0.5x². Let me check when R(x) is positive.R(x) = 500x - 0.5x² > 0Factor out x:x(500 - 0.5x) > 0So, critical points at x=0 and x=1000.Since x is positive, the revenue is positive when 0 < x < 1000.So, our previous solution x between 21 and 407.57 is within the domain where revenue is positive.But wait, let me double-check the inequality.We had:0.35x² - 150x + 3000 ≤ 0Which is true for x between 21 and 407.57. So, the profit margin is at least 30% when x is between approximately 21 and 407.57 units.But since the number of units must be an integer, we can say x is from 21 to 407 units.But let me verify this with an example. Let's pick x=21:Compute P(21) and R(21):P(21) = -0.5*(21)^2 + 300*21 - 3000= -0.5*441 + 6300 - 3000= -220.5 + 6300 - 3000= (6300 - 3000) - 220.5= 3300 - 220.5 = 3079.5R(21) = 500*21 - 0.5*(21)^2= 10500 - 0.5*441 = 10500 - 220.5 = 10279.5Profit margin = 3079.5 / 10279.5 ≈ 0.3, which is 30%. So, that's the lower bound.Similarly, at x=407.57, let's compute P(x)/R(x):But since x must be integer, let's check x=407 and x=408.First, x=407:P(407) = -0.5*(407)^2 + 300*407 - 3000Compute 407²: 407*407. Let's compute 400²=160000, 2*400*7=5600, 7²=49. So, (400+7)^2=160000 + 5600 +49=165649So, P(407) = -0.5*165649 + 300*407 - 3000= -82824.5 + 122100 - 3000= (-82824.5 - 3000) + 122100= -85824.5 + 122100 ≈ 36275.5R(407) = 500*407 - 0.5*(407)^2= 203500 - 0.5*165649 ≈ 203500 - 82824.5 ≈ 120675.5Profit margin = 36275.5 / 120675.5 ≈ 0.3, which is 30%.Similarly, x=408:P(408) = -0.5*(408)^2 + 300*408 - 3000408² = 166464So, P(408) = -0.5*166464 + 122400 - 3000= -83232 + 122400 - 3000= (-83232 - 3000) + 122400 ≈ -86232 + 122400 ≈ 36168R(408) = 500*408 - 0.5*(408)^2= 204000 - 0.5*166464 ≈ 204000 - 83232 ≈ 120768Profit margin = 36168 / 120768 ≈ 0.2995, which is approximately 29.95%, just below 30%.So, x=408 gives just below 30%, so the upper limit is x=407.Similarly, at x=21, it's exactly 30%, and below that, say x=20:P(20) = -0.5*(400) + 300*20 -3000 = -200 + 6000 -3000 = 2800R(20) = 500*20 -0.5*(400) = 10000 -200=9800Profit margin = 2800 /9800 ≈ 0.2857, which is 28.57%, below 30%. So, x must be at least 21.Therefore, the range of x is from 21 to 407 units.Wait, but let me just confirm if the quadratic inequality solution was correct. We had:0.35x² - 150x + 3000 ≤ 0Which is true between the roots 21 and 407.57. So, x must be between 21 and 407.57. Since x must be an integer, x=21 to x=407.But let's also check x=0, just to make sure. At x=0, profit is -3000, revenue is 0, so profit margin is undefined. So, x must be greater than 0.Also, at x=1000, revenue is 0, so profit margin is undefined again.So, the valid range is x from 21 to 407.Therefore, the answers are:1. The number of units that maximizes profit is 300.2. The range of units that gives a minimum profit margin of 30% is from 21 to 407 units.I think that's it. Let me just recap:For part 1, we found the profit function, took its derivative, found the critical point, confirmed it's a maximum with the second derivative.For part 2, we set up the profit margin inequality, solved the quadratic inequality, found the bounds, and checked the endpoints to ensure they satisfy the condition.Everything seems to check out. I don't think I made any calculation errors, but let me just quickly verify one more thing.In part 2, when I multiplied both sides by -1, I correctly reversed the inequality. So, that step was correct.Also, when solving the quadratic equation, I correctly applied the quadratic formula, calculated the discriminant, approximated the square root, and found the roots. Then, since the parabola opens upwards, the inequality is satisfied between the roots.Yes, that all makes sense.So, I think I'm confident with these answers.Final Answer1. The number of units that maximizes profit is boxed{300}.2. The range of units that satisfies the minimum profit margin is boxed{[21, 407]}.</think>