Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functionsc) We know that the demand is x = 60 when the revenue is maximum. Our task is tofind what p is. Use p = (−1/3)x+40 and plug in 60 for x to get p = (−1/3)60+40 =−20 + 40 = 20. The unit price should be $20 dollars to yield maximum revenue.d) The maximum revenue itself is the R-coordinate of the vertex. We already havex = 60. Plug this into the equation for R to get R(60) = (−1/3)60 2 + 40(60) =$1200.39. The dimensions of the rectangle are x by y, so its area is A = xy. We do notknow how to find maximums of equations with more than one variable, so we need toget this down to an equation for A in terms of a single variable. Luckily, we are giventhat y = mx + b, so replace y with mx + b to get A = x(mx + b). Multiply this outto get A = mx 2 + bx, which is a quadratic equation. We are given that m < 0, so thegraph of the area function A is a downward parabola, meaning its maximum occurs atthe vertex. Use x = −b/(2a) = −b/(2m) to get one dimension. Now plug in to get( ) −by = mx + b = m + b = − b 2m 2 + b = − b 2 + 2b2 = b 2 ,andA = xy =( ( ) −b b= −2m) b22 4m .So, the maximum area of −b 2 /(4m) occurs when the dimensions are −b/(2m) by b/2.Version: Fall 2007