cwhat is the revenue if 200 units are sold round to the near

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.)

Solution

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