c)what is the revenue if 200 units are sold? (round to the nearest hundreth as n

Question

c)what is the revenue if 200 units are sold? (round to the nearest hundreth as needed)

d) what quanity x maximizes revenue?

what is the maximum revenue?

e) what price should the company charge to maximize revenue?

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation p x+ 100. Answer parts (a) through (e). (a) Find a model that expresses the revenue R as a function of x. (Remember, R- xp.) R(x) (Use integers or fractions for any numbers in the expression.)

Explanation / Answer

a)
R(x) = x*p
= x*(-1/9 *x + 100)
= (-1/9)*x^2 + 100*x

b)
x is quantity and it can't be negative
so, x>=0

c)
R(x) = (-1/9)*x^2 + 100*x
put x=200,
R(x) = (-1/9)*200^2 + 100*200
= 15556
Rounding we get,
Revenue = 15600

d)
R(x) = (-1/9)*x^2 + 100*x
differentiate with respect to x,
R'(x) = (-2/9)*x +100
put R'(x) = 0
(-2/9)*x +100 = 0
x = 450

for x>450, R'(x) is negative and for x<450, R'(x) is positive

so, x = 450 maximises revenue

maximum revenue = R(450)
= (-1/9)*450^2 + 100*450
= 22500

e)
x = 450
p = (-1/9)*x +100
= (-1/9)*450 + 100
= 50

price that should be charged = 50

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