please help me with this precal question!?

Question

Demand Equation, R= XP

the price p and the quantity x sold of a certain product obey the demand equation

x= -5p + 100, 0 less than or equal to (p) less than or equal to 20

(a) express the revenue r as a function of x.
(b) what is the revenue if 15 units are sold?
(c) what quantity x maximizes revenue? what is the maximum revenue?
(d) what price should the company charge to maximize revenue?

I DONT UNDERSTAND THIS STUFF VERY WELL SO PLEASE BE DETAILED.

i dont know how to take the derivative, dose it work with completeing the square?

Answer 1

Let
x = number of units sold
p = price per unit
R = revenue

Given
x = -5p + 100, where 0 ≤ p ≤ 20

We have
x = -5p + 100
5p = -x + 100
p = -x/5 + 20

(a)
R = xp = x(-x/5 + 20) = -x²/5 + 20x
R(x) = -x²/5 + 20x
______________________

(b)
R(x) = -x²/5 + 20x
R(15) = -(15)²/5 + 20(15) = -225/5 + 300 = -45 + 300 = 255
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(c)
Take the derivative and set equal to zero to find the critical points.

dR/dx = -2x/5 + 20 = 0
2x/5 = 20
2x = 100
x = 50

Take the second derivative to find the nature of the critical point.

d²R/dx² = -2/5 < 0
This implies a relative maximum which is what we want.

R(x) = -x²/5 + 20x
R(50) = -(50)²/5 + 20(50) = -500 + 1000 = 500

(c)
The number of units that maximizes revenue is 50. Maximum revenue is 500.
________________________________

(d)
p = -x/5 + 20 = -50/5 + 20 = -10 + 20 = 10
The constraint 0 ≤ p ≤ 20 holds, so it is a valid solution.

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Solving without calculus.

As an aside, on (c) where we are asked to maximize revenue. You want to find the vertex of the parabola which is concave downward.

R(x) = -x²/5 + 20x = (-1/5)(x² - 100x + 2500) + 500
R(x) - 500 = (-1/5)(x - 50)²

So maximum revenue is 500 when 50 units are sold.

So you are correct. You want to complete the square.

Answer 2

p = r/x
p = (100 - x)/5
r/x = (100 - x)/5
r = x(100 - x)/5
r(15) = 15(100 - 15)/5
r(15) = 3(85)
r(15) = 255
To find the maximum r without calculus, put the equation into vertex form:
r = -(1/5)(x^2 - 100x)
r = -(1/5)(x^2 - 100x + 2,500 - 2,500)
r = -(1/5)(x - 50)^2 + 500
r - 500 = -(1/5)(x - 50)^2
We find that the revenue is a maximum of 500 when x = 50
From the 1st equation above, p = 500/50 = 10

Answer 3

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