1.) triangular corner lot has perpendicular sides of lengths 120 feet and 160 feet. Find the dimensions of the largest rectangular building that can be constructed on the lot with sides parallel to the streets.

Answer 1

let the length of the rectangle parallel to the 160 ft side be x
let its height be y
I see 3 similar triangles, the original and the 2 smaller triangles
by ratios:
y/(160-x) = 120/160 = 3/4
4y = 480 - 3x
y = 120 - 3x/4

area of rectangle = A
= xy
= x(120 - 3x/4) = 120x - 3x^2/4
dA/dx = 120 - 3x/2
= 0 for a max of A
120 - 3x/2 = 0
3x = 240
x = 80 , as expected
then y = 120 - 3(80)/4 = 60 , also as expected

the rectangle should be 80 ft by 60 ft , (the midpoints of the sides)