Asked by I

A rectangle has a vertex on the line 3x + 4y = 12.
Two of its sides lie along the positive x and y axes.
Find the maximum possible area of such a rectangle.
Answered by Reiny
Let the vertex on the line 3x + 4y = 12 be (x,y)
then y = (12-3x)/4 = 3- 3x/4

area = xy = x(3 - 3x/4) = 3x - 3x^2 /4
d(area)/dx = 3 - (3/2)x
= 0 for a max of area
(3/2)x = 3
3x = 6
x = 2 , y = 3 - 6/4 = 3/2

max area = xy = 2(3/2) = 3 units^2

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