A producer find that demand for his commodity obeys a linear demand equation p+0.85x=8.5, where p is in thousands of dollars and x is in thousands of units. If the producer's cost are given by C(x)=1+4x, what should his level of production be to maximize profits?

the level of production is=

(in thousands of units)

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First we must determine the equation for profit. This will be equal to the total revenue subtracted by the production costs:

`P(x) = R(x) - C(x)`

We know that C(x) = 1+4x, but we need to solve for R(X). This will be equal to the total number of units sold multiplied by the cost of those units:

`R(x) = p(x)x`

`p(x) = 8.5 - 0.85x`

Therefore:

`R(x) = (8.5-0.85x)x = 8.5x-0.85x^2`

Thus:

`P(x)=8.5x-0.85x^2-(1+4x) = -0.85x^2+4.5x-1`

In order to find the maximum profits we must find for what value of x the derivative (change in profits per change in x) of P(x) is equal to 0:

`(dP)/dx=-1.7x+4.5=0 -> x=2.6`

Therefore, if approximately 2600 units are sold, profits will be at a maximum

Let us say cost function is C(x), revenue function is R(x) and profit function P(x).

Profit = Revenue-Cost

`P(x) = R(X)-C(x)`

Cost of x thousand items sold;

`C(x) = (1+4x)`

Revenue of x thousand items sold;`R(x) = (8.5-0.85x)x = 8.5x-0.85x^2`

`P(x) = (8.5x-0.85x^2)-(1+4x)`

`P(x) = -0.85x^2+4.5x-1`

For maximum or minimum profit P'(x) = 0

`P'(x) = -2xx0.85x+4.5 = -1.7x+4.5`

When P'(x) = 0;

`0 = -1.7x+4.5`

`x = 2.647`

By second derivative test, if P(x) has a maximum at x = 2.647 then at the same point P''(x) should be negative.

`P'(x) = -1.7x+4.5`

`P''(x) = -1.7`

`P''(x)<0`

So P(x) has a maximum at x = 2.647

*So the maximum profit will be generated when the sold items comes to a value of 2647. Or in thousand units it should be 3000 (x = 3).*

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