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