find the dimensions of the rectangular area of maximum area which can be laid out within a triangle of base 12 and altitude 4 if one side of the rectangle lies on the base of the triangle

Question

thanks

Answer 1

The height of the rectangle can be anything from 0 to 4. Call it h.

The width (4) of the rectangle varies linearly from 12 to 0, with
w = 12 (1 - h/4)= 12 - 3h

Area = f(h) = h*w = 12h - 3h^2

dA/dx = 0 when 6h = 12
h = 2; w = 12 - 6 = 6
Amax = 12

The triangle does not have to be isosceles

Answer 2

That is a problem you do when refreshed, and have some time.

Here it is worked given the triangle vertexes.

You are given two point, and the altidude. For your upper vertex, x,y, choose it such that the altitude (y) is 4.

http://www.analyzemath.com/calculus/Problems/maximize_area_rectangle.html

Yours is somewhat more difficult.