Find the maximum value of xy if 3x+4y=12

Answer 1

3x+4y = 12

y = (12-3x)/4 = 3 -3x/4

xy = x(3-3x/4) = 3x -3x^2/4

A' = (xy)' = 3 - 3x/2 = 0

1-x/2 = 0

x/2 = 1

x =2

y = 3-6/4 = 3/2

x = 2, y= 3/2

max Area = xy = 2(3/2) = 3

another way to solve the problem is graph the equation

and find the midpoint of the line segment connecting the intercepts

(0,3), (4,0) which is (2, 3/2) then multiply the x and y coordinates

max xy = (2)(3/2) = 3