college algebra

Question

the price p and the quantity X sold of a small flatscreen television set obeys the demand equation below

p=-.16x+320

a) how much should be charged for the television set if there are 90 television sets in stock?

b) what quantity X will maximize revenue? What is the maximum revenue?

c) what price should be charged in order to maximize revenue ?

Answer 1

a) P(x) = -.16x +320 (the given demand equation)

plug in 90 for x:

P(90) = -.16 (90) +320 = $305.60 per unit

b) and c) Note that Revenue is equal to $price/unit times the number of units, in math terms:

R = Px

Now, take the given demand equation and solve for x:

P = -.16x + 320

P - 320 = -.16x

(P - 320)/-.16 = x =>>> x = 2000 - 6.25P

Now, substitute this equation for x into the Revenue equation (R = Px), which gives us:

R = Px = P (2000 - 6.25P), which we then multiply the terms to give us the quadratic equation, which is the general equation for a parabola:

R = -6.25P² +2000P

For a quadratic equation, the maximum R will occur where x = -b/2a (this is the parabola's vertex, it's highest point)

Note: a = -6.25 and b = 2000, therefore -b/2a = -2000/2(-6.25) = 160, which means

the maximum Revenue occurs at x = 160 units.

Plugging in 160 for x in the original given demand equation, gives us

P(160) = -.16 (160) +320 = $294.40 per unit. Plug in this price into the Revenue equation R = Px

R_max = $294.40/unit x 160 units = $47,104 total revenue