Optimization Problems

Question

A rectangle is bounded by the x- and y-axes and the graph of  y = (6 − x)/2 (see figure). What length (x) and width (y) should the rectangle have so that its area is a maximum?

Answer 1

y=(6-x)/2 = -x/2 + 3, slope = -1/2, y intercept = (0,3)

x intercept = (6,0)

the midpoint of the line between the x and y intercepts is ((0+6)/2, (0+3)/2) = (3, 3/2)

that's the vertex point that maximizes the area of the rectangle = 3(3/2) = 9/2 = 4.5 square units

this problem is similar to a linear demand curve, where you try to find the revenue maximizing price and quantity sold. It's the midpoint of the demand cuve. midway between the 2 intercepts.

or use calculus

A=xy=x(6-x)/2= 3x-x^2/2

A'=3-x=0

x=3

y=(6-3)/2=3/2

A=xy=3(3)-3^2/2=9/2