p ( x ) = - 1 15 x + 30 ;  0 ≤ x ≤ 450 write an equation for the revenue are if the revenue is the price times the number of units sold. what price should a company charge to have maximum revenue

Answer 1

If you haven’t learned about derivatives yet, you can use the fact that quadratic equations form parabolas. A parabola has a minimum if its leading coefficient is positive and a maximum if the leading coefficient is negative. Moreover, this value occurs at the vertex (h,k). If the quadratic equation is in the form ax² + bx + c = 0, then the vertex occurs at h=-b/(2a). Take this value substitute it into p(x) and you’ll have your answer. I didn’t work anything out here because I think you might have mistyped the problem, but the revenue function is r(x)=x*p(x). This will form the quadratic equation to utilize this process on.

Answer 2

Revenue = P time x = (-150x+30)x = -150x^2 +30x

take the derivative and set = 0

-300x + 30 = 0

x=30/300 = 1/10

Maximum revenue = -150(1/10)^2 + 30(1/10) = -150/100 + 30/10 = -3/2 + 3 = 3/2

revenue maximizing price = -150(1/10)+30 = -15+30 = 15

this doesn't look right. Maybe the price function P(x)=-1/15x + 30 ? did you leave out a /, division sign?

Px then = -1/15 +30x.

then you maximize Revenue by maximizing quantity sold. x=450

then revenue = 30(450)-1/15 = 1350 - 1/15 = 1349 14/15

price and quantity sold are inversely related.

either price function

P(x)=-115x+30 or P(x)=-1/15x+30 is an inverse relation, although the 2nd makes more economic sense

The first is linear, the 2nd is hyperbolic