The Cost To Produce X Thousand Automobiles Is Given By The Function?
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c(x) = 4×2 – 8x + 3.
The price that we will charge is
p(x) = 16 – 2x.
What is the revenue? What is the profit for producing x thousand automobiles? What is the number of automobiles that must be produced to maximize the profit?
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revenue: R(x) = x*p(x) = 16x – 2x^2
profit: Pr(x) = R(x) – c(x) = (16x – 2x^2) – (4x^2 – 8x + 3) = -6x^2 + 24x – 3
if you’re using calculus, then the profit is maximized when d/dx (Pr(x)) = 0, so
-12x + 24 = 0
12x = 24
x=2
If you’re not using calculus, then you want the marginal revenue to equal the marginal cost… but i forget how to calculate those without calculus.
Let F = profit.
Then F = total revenue – cost
F = p(x) – c(x) = 13+6x-4x^2
When x=0, F = 13,000.
When x=1, F = (13 +6-4)thousand = 15,000.
thereafter the profit decreases as x increases.
To find turning point, differentiate function and equate to zero. dF/dx = 6-8x. dF/dx = 0 gives x=3/4.
When x=3/4, F = 13 + 9/4 = 61/4 thousand
F = 15,250.
Number of automobiles to maximise profit is 15,250.