If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where p(x) = 39 - . How many candy bars must be sold to maximize revenue?
I thought I could solve this problem by using implicit differentiation
but don't know how to... Please help.
Ok.
p(x) = 39 - x/24 is the price for a candy bar
you sell x candy bars, then your profit is
x*p(x) = 39x - x^2 / 24 =: f(x)
=> f'(x) = 39 - 2*x^1 / 24 = 39 - x/12
(f''(x) = - 1 /12 < 0 => Maximum )
f'(x) = 0
39 - x/12 = 0
// + x/12
39 = x/12
// * 12
x = 39 * 12 = 468
f''(x) < 0 => x = 468 Maximum
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x*p(x) = 39x - x^2 / 24 =: f(x)
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f(468) = 468*(39 - 468 / 24) = 9126
Im still confused about
p(x) is the price charged for ONE candy bar, p(x) = 39 - x /24,
but x thousand candy bars will be sold
not sure about, maybe it is x*p(x)*1000, I dont know