Question
What is the maximum area of a rectangle that has sides along the positive x-axis and y-axis and lies below the line 3x+2y=1
Answers
Reiny
Let the right-hand corner vertex be P(x,y)
but 3x+2y = 1
y = (1-3x)/2 or 1/2 - (3/2)x
area = xy
= x(1/2 - (3/2)x)
= (1/2)x - (3/2)x^2
d(area)/dx = 1/2 - 3x
= 0 for a max of area
3x = 1/2
x = 1/6
y = 1/2 - (3/2)(1/6) = 1/4 ---> P is (1/6 , 1/4)
max area = xy = (1/6)(1/4) = 1/24 square units
but 3x+2y = 1
y = (1-3x)/2 or 1/2 - (3/2)x
area = xy
= x(1/2 - (3/2)x)
= (1/2)x - (3/2)x^2
d(area)/dx = 1/2 - 3x
= 0 for a max of area
3x = 1/2
x = 1/6
y = 1/2 - (3/2)(1/6) = 1/4 ---> P is (1/6 , 1/4)
max area = xy = (1/6)(1/4) = 1/24 square units