Math, asked by sudha060982, 4 months ago

47. A steel plant is capable of producing x tonnes per day of a low - grade steel and y tonnes
40-5x
per day of a high-grade steel , where y = If the fixed market price of low - grade steel
10-X
is half that of high-grade steel, then what should be optimal productions in low - grade steel
and high-grade steel in order to have maximum receipts. (OR)​

Answers

Answered by sindhusjamyal
2

Answer:

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Step-by-step explanation:

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Answered by Rameshjangid
0

Answer:

In summary, to maximize revenue, the steel plant should produce 150 tonnes per day of low-grade steel and zero tonnes per day of high-grade steel.

Explanation:

To find the optimal productions of low-grade and high-grade steel, we need to maximize the total revenue. Let's first determine the revenue function:

Revenue = (price of low-grade steel)(quantity of low-grade steel) + (price of high-grade steel)(quantity of high-grade steel)

Let p be the price of high-grade steel. Then the price of low-grade steel is 0.5p. We can now substitute the given expressions for x and y into the revenue function:

Revenue = (0.5p)(x) + (p)(40-5x)

Simplifying the expression:

Revenue = 40p - 1.5px

Now, we need to find the values of x and y that maximize the revenue, subject to the constraints given:

y = 40 - 5x (equation 1)

0.5p = price of low-grade steel

p = price of high-grade steel

We can substitute equation 1 into the revenue equation to eliminate y:

Revenue = (0.5p)(x) + (p)(40-5x)

Revenue = 20p - 0.5px

We can now rewrite the problem as:

Maximize 20p - 0.5px subject to x≥0 and p≥0

To solve this problem, we need to take partial derivatives with respect to x and p, and set them equal to zero:

∂(20p - 0.5px)/∂x = -0.5p = 0

∂(20p - 0.5px)/∂p = 20 - 0.5x = 0

Solving for p and x:

p = 0

x = 40

Since p = 0, the price of high-grade steel is zero, and we should only produce low-grade steel. Therefore, the optimal production of low-grade steel is 150 tonnes per day (when x=40, y=0), and the optimal production of high-grade steel is zero.

In summary, to maximize revenue, the steel plant should produce 150 tonnes per day of low-grade steel and zero tonnes per day of high-grade steel.

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