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Give step-by-step solution with explanation and final answer:Afarm produces cheese and sells it by weight. ’ a The corresponding revenue function® is RG) = 91x and the daily cost function® is C(x) = 45 + 127x088 where R(x) and C(x) are in euros and x denotes the weight of cheese sold, in kilograms. For practical reasons (milk supply, storage and display space), the maximum amount of cheese that the farm can produce and sell daily is 31.1 kilograms. 1) Find an interval with length at most 10 grams that contains a break-even® point for the sales of cheese, where the break-even point is expressed in kilograms. 2) Prove that the sale of cheese will produce a maximum profit. Hence, find how many kilograms of cheese should be sold to maximise profit. In addition, in order to avoid the worst possible financial situation, the farmers would also like to know selling how many kilograms of cheese will result in the greatest loss. In order to answer this question, find how many kilograms of cheese sold will minimise profit’. The revenue function of a product describes the amount of money generated by the sales of a certain quantity of the product. It is obtained by multiplying the quantity of product sold by the price function. For instance, the most common revenue function is y = bx, where y is the total revenue, b is the selling price per unit of sales, and x is the number of units sold. Note that the revenue function does not take into account the costs (but the profit does, see below). The cost function of a product measures the amount of money required to be spent to sell a certain amount of the product. It is obtained by adding the fixed cost (for instance: store rental, utility bills), and the variable cost (for instance: the supplies needed to make the product). ©The break-even point occurs when the revenue is equal to the cost of a product. When a company breaks even, it neither wins nor loses money as the revenue exactly balances the cost associated to that product. 9 Profitis the difference between Revenue and Cost, namely P(x) = R(x) — C(x).A negative profit corresponds to a financial loss. The questions below his line are for you to make some checks to ensure you are on the right track with your calculations and explanations. 1) 8) The theorem we will be using for question 1.a) is. he maxcin theorom ho mermediato value theorem ne mean valu theorem b) Enter your interval using standard notation with round or square brackets e.g. (a,b) of [a, b] : The break-even point x, expressed in kilograms, is in the following interval of length at most 10 grams |: J] Hints for writing your solution: State the assumptions of the theorem you are planning to use, and verify that they hold true. 2) a) The theorem we will be using for question 2) is he main theorem no nermedate vale theorem 10 mean valu theorem “The weight of cheese which should be sold fo maximise profits Number Kiograms, rounded to the nearest gram. The sale of cheese which wil result in the lowest profit is Number kilograms, rounded to the nearest gram. Hints for writing your solution: i. State the assumptions of the max-min theorem, and verify that they hold true on the interval [0, 31.1]. i Find 92. where P i he prof. iii. Remember that we want to find the values of x that correspond respectively to the maximum of and minimum of P.

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Give step-by-step solution with explanation and final answer:Uploaded ImageUploaded ImageAfarm produces cheese and sells it by weight. ’ a The corresponding revenue function® is RG) = 91x and the daily cost function® is C(x) = 45 + 127x088 where R(x) and C(x) are in euros and x denotes the weight of cheese sold, in kilograms. For practical reasons (milk supply, storage and display space), the maximum amount of cheese that the farm can produce and sell daily is 31.1 kilograms. 1) Find an interval with length at most 10 grams that contains a break-even® point for the sales of cheese, where the break-even point is expressed in kilograms. 2) Prove that the sale of cheese will produce a maximum profit. Hence, find how many kilograms of cheese should be sold to maximise profit. In addition, in order to avoid the worst possible financial situation, the farmers would also like to know selling how many kilograms of cheese will result in the greatest loss. In order to answer this question, find how many kilograms of cheese sold will minimise profit’. The revenue function of a product describes the amount of money generated by the sales of a certain quantity of the product. It is obtained by multiplying the quantity of product sold by the price function. For instance, the most common revenue function is y = bx, where y is the total revenue, b is the selling price per unit of sales, and x is the number of units sold. Note that the revenue function does not take into account the costs (but the profit does, see below). The cost function of a product measures the amount of money required to be spent to sell a certain amount of the product. It is obtained by adding the fixed cost (for instance: store rental, utility bills), and the variable cost (for instance: the supplies needed to make the product). ©The break-even point occurs when the revenue is equal to the cost of a product. When a company breaks even, it neither wins nor loses money as the revenue exactly balances the cost associated to that product. 9 Profitis the difference between Revenue and Cost, namely P(x) = R(x) — C(x).A negative profit corresponds to a financial loss. The questions below his line are for you to make some checks to ensure you are on the right track with your calculations and explanations. 1) 8) The theorem we will be using for question 1.a) is. he maxcin theorom ho mermediato value theorem ne mean valu theorem b) Enter your interval using standard notation with round or square brackets e.g. (a,b) of [a, b] : The break-even point x, expressed in kilograms, is in the following interval of length at most 10 grams |: J] Hints for writing your solution: State the assumptions of the theorem you are planning to use, and verify that they hold true. 2) a) The theorem we will be using for question 2) is he main theorem no nermedate vale theorem 10 mean valu theorem “The weight of cheese which should be sold fo maximise profits Number Kiograms, rounded to the nearest gram. The sale of cheese which wil result in the lowest profit is Number kilograms, rounded to the nearest gram. Hints for writing your solution: i. State the assumptions of the max-min theorem, and verify that they hold true on the interval [0, 31.1]. i Find 92. where P i he prof. iii. Remember that we want to find the values of x that correspond respectively to the maximum of and minimum of P.

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Asked by: gvr
Created at: 2025-06-19 23:40:50

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Let's break down and solve the questions step by step. ### Given: - **Revenue function:** \( R(x) = 91x \) - **Cost function:** \( C(x) = 45 + 127x^{0.88} \) - Maximum cheese produced daily: \( 31.1 \) kg --- ## **1. Finding the break-even point** **Break-even:** \( R(x) = C(x) \) So, \[ 91x = 45 + 127x^{0.88} \] \[ 91x - 127x^{0.88} - 45 = 0 \] Let's define \( f(x) = 91x - 127x^{0.88} - 45 \). ### 1a) **The theorem we'll use** Since \( f(x) \) is continuous and we are looking for a root in a closed interval (where \( f(a) \) and \( f(b) \) have opposite signs), **the intermediate value theorem** applies. **Answer:** - [x] the intermediate value theorem --- ### 1b) **Find an interval of length at most 10 grams (0.01 kg) containing the break-even point** Let's test values (in kg): - At \( x = 0 \): \[ f(0) = 91(0) - 127(0)^{0.88} - 45 = -45 \] - At \( x = 1 \): \[ f(1) = 91(1) - 127(1)^{0.88} - 45 = 91 - 127 - 45 = -81 \] - At \( x = 2 \): \[ f(2) = 91(2) - 127(2)^{0.88} - 45 \\ 2^{0.88} \approx 1.845 \\ 91 \times 2 = 182 \\ 127 \times 1.845 \approx 234.415 \\ f(2) = 182 - 234.415 - 45 \approx -97.415 \] - At \( x = 5 \): \[ 5^{0.88} \approx 4.123 \\ 127 \times 4.123 \approx 523.621 \\ 91 \times 5 = 455 \\ f(5) = 455 - 523.621 - 45 \approx -113.621 \] - At \( x = 15 \): \[ 15^{0.88} \approx 11.663 \\ 127 \times 11.663 \approx 1,480.201 \\ 91 \times 15 = 1,365 \\ f(15) = 1,365 - 1,480.201 - 45 \approx -160.201 \] - At \( x = 20 \): \[ 20^{0.88} \approx 15.832 \\ 127 \times 15.832 \approx 2,011.864 \\ 91 \times 20 = 1,820 \\ f(20) = 1,820 - 2,011.864 - 45 \approx -236.864 \] - At \( x = 30 \): \[ 30^{0.88} \approx 22.641 \\ 127 \times 22.641 \approx 2,875.407 \\ 91 \times 30 = 2,730 \\ f(30) = 2,730 - 2,875.407 - 45 \approx -190.407 \] It looks like the function remains negative. Let's try higher values for \( x \): Wait, this shouldn't be the case. Let's check at \( x = 0.5 \): \[ 0.5^{0.88} \approx 0.541 \\ 127 \times 0.541 \approx 68.707 \\ 91 \times 0.5 = 45.5 \\ f(0.5) = 45.5 - 68.707 - 45 = -68.207 \] Try \( x = 1.5 \): \[ 1.5^{0.88} \approx 1.406 \\ 127 \times 1.406 \approx 178.562 \\ 91 \times 1.5 = 136.5 \\ f(1.5) = 136.5 - 178.562 - 45 = -87.062 \] Try \( x = 5 \): \[ 5^{0.88} \approx 4.123 \\ 127 \times 4.123 \approx 523.621 \\ 91 \times 5 = 455 \\ f(5) = 455 - 523.621 - 45 \approx -113.621 \] Try \( x = 10 \): \[ 10^{0.88} \approx 7.585 \\ 127 \times 7.585 \approx 963.295 \\ 91 \times 10 = 910 \\ f(10) = 910 - 963.295 - 45 = -98.295 \] Try \( x = 20 \): \[ 20^{0.88} \approx 15.832 \\ 127 \times 15.832 \approx 2011.864 \\ 91 \times 20 = 1820 \\ f(20) = 1820 - 2011.864 - 45 = -236.864 \] Try \( x = 31.1 \): \[ 31.1^{0.88} \approx 25.045 \\ 127 \times 25.045 \approx 3180.715 \\ 91 \times 31.1 = 2830.1 \\ f(31.1) = 2830.1 - 3180.715 - 45 = -395.6 \] All values are negative, suggesting a possible calculation error or misunderstanding. Let's try to solve for \( x \) algebraically: \[ 91x - 127x^{0.88} = 45 \] Let's plot or estimate more closely. Let's try \( x = 0.6 \): \[ 0.6^{0.88} \approx 0.655 \\ 127 \times 0.655 = 83.185 \\ 91 \times 0.6 = 54.6 \\ f(0.6) = 54.6 - 83.185 - 45 = -73.585 \] Try a much higher \( x \), say \( x = 50 \): \[ 50^{0.88} \approx 33.74 \\ 127 \times 33.74 = 4285.98 \\ 91 \times 50 = 4550 \\ f(50) = 4550 - 4285.98 - 45 = 219.02 \] Now, at \( x = 40 \): \[ 40^{0.88} \approx 28.45 \\ 127 \times 28.45 = 3613.15 \\ 91 \times 40 = 3640 \\ f(40) = 3640 - 3613.15 - 45 = -18.15 \] So, between \( x = 40 \) and \( x = 50 \), \( f(x) \) goes from negative to positive. But our maximum daily production is \( x = 31.1 \) kg! So within this range, \( f(x) < 0 \) always. **Conclusion: There is no break-even point in the interval [0, 31.1], as \( f(x) \) is always negative.** ### Check for errors: Let’s check calculations for \( x = 31.1 \): \[ 31.1^{0.88} \approx 25.045 \\ 127 \times 25.045 = 3180.715 \\ 91 \times 31.1 = 2830.1 \\ f(31.1) = 2830.1 - 3180.715 - 45 = -395.6 \] So, the cost always exceeds the revenue in the feasible range. **No break-even point exists in the interval [0, 31.1].** --- ## **2. Maximising and Minimising Profit** ### Define profit: \[ P(x) = R(x) - C(x) = 91x - (45 + 127x^{0.88}) = 91x - 45 - 127x^{0.88} \] #### 2a) The theorem we'll use: Here, since \( P(x) \) is continuous on a closed interval, and we are seeking the maximum and minimum, we use the **max-min theorem**. **Answer:** - [x] the max-min theorem --- #### Find where profit is maximised/minimised Set derivative \( P'(x) = 0 \): \[ P'(x) = 91 - 127 \times 0.88 x^{-0.12} \] \[ = 91 - 111.76x^{-0.12} \] Set to zero and solve: \[ 91 = 111.76x^{-0.12} \] \[ \frac{91}{111.76} = x^{-0.12} \] \[ \ln \left( \frac{91}{111.76} \right) = -0.12 \ln x \] \[ \ln x = -\frac{\ln \left( \frac{91}{111.76} \right)}{0.12} \] Calculate: \[ \frac{91}{111.76} \approx 0.8146 \] \[ \ln(0.8146) \approx -0.205 \] \[ \ln x = - \frac{-0.205}{0.12} \approx 1.708 \] \[ x = e^{1.708} \approx 5.52 \] So, the profit is maximised at \( x \approx 5.52 \) kg. --- #### Check endpoints for min profit: - \( x = 0 \): \[ P(0) = 91 \times 0 - 45 - 127 \times 0^{0.88} = -45 \] - \( x = 31.1 \): \[ 31.1^{0.88} \approx 25.045 \\ 127 \times 25.045 = 3180.715 \\ 91 \times 31.1 = 2830.1 \\ P(31.1) = 2830.1 - 45 - 3180.715 = -395.615 \] So, the **minimum profit** occurs at the endpoint \( x = 31.1 \) kg. --- ## **Summary of Answers** ### 1a) Theorem: - **Intermediate Value Theorem** ### 1b) Interval: - **There is no break-even point in [0, 31.1], as profit is always negative (cost always exceeds revenue).** ### 2a) Theorem: - **Max-min theorem** ### 2b) Maximum profit: - **5.52 kg** (rounded: **5.52 kg**) ### 2c) Minimum profit: - **31.1 kg** --- **If you need interval notation for a break-even point, write:** > No break-even interval exists in [0, 31.1], as cost always exceeds revenue. --- **Final Answers:** - **1a:** Intermediate value theorem - **1b:** No interval, as there is no break-even point in [0, 31.1] - **2a:** Max-min theorem - **Profit is maximised at:** **5.52 kg** - **Profit is minimised at:** **31.10 kg**

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