Business Mathematics

[mathdad]

The price p and the quantity x sold of a certain product obey the demand equation below.
x= -8p + 144, 0 ≤ p ≤ 18

(a) Express the revenue R as a function of x.

Solution:

I need R(x).

Revenue = (price)(quantity)

R = xp

I gotta solve the demand equation for p.

x = -8p + 144

(x - 144) = -8p

(x - 144)/-8 = p

(-1/8)(x - 144) = p

(-x/8) + 18 = p

R(x) = xp

R(x) = x[(-x/8) + 18]

R(x) = (-x^2)/8 + 18x

(b) What is the revenue if 88 units are sold?

Solution:

Let x = 18

R(18) = -(18)^2/8 + 18(18)

R(18) = 283.5 dollars

Can revenue be a decimal number?

(c) What quantity x maximizes revenue? What is the maximum revenue?

Solution:

R(x) = (-x^2)/8 + 18x

x = -b/2a

x = -(18)/2(-1/8)

x = -18/(-1/4)

x = 18/(1/4)

x = 18 • 4

x = 72

R(72) = max revenue

R(72) = -(72)^2)/8 + 18(72)

R(72) = 648 dollars

Before continuing to do parts d and e, is everything correct so far?

(d) What price should the company charge to maximize revenue?

(e) What price should the company charge to earn at least $520 in revenue?

 

[Steven G]

It looks OK.

 

[mathdad]

It looks OK.

Thank you. The rest will be solved on paper.