Find the rectangle with largest area?

Question

Find the rectangle with largest area which can be inscribed in a right triangle with sides

5 and 12 cm such that two adjacent sides of the rectangle lie on legs of the triangle.

Answer 1

Let the side of the rectangle on the 5cm leg be x, and the other side be y.

By similar triangle, y/12=(5-x)/5

=> y=12(5-x)/5

Area of rectangle, A = xy

= 12(5x-x^2)/5

A'=12(5-2x)/5

A'=0 => x=5/2

y=6

Check: A''=12(-2)/5<0 => max point

Answer 2

The area is going to x*y.  Now to find an equation for y in terms of x that bounds it to the hypotenouse of the triangle.  If we assume the triangle starts off at (0,12) and ends up at (5,0) in cm, then we can write an equation for y interms of x from the straight line equation.

y = mx + b,

y = mx + 12

0 = 5m +12

m = -12/5

so, y = -(12/5)x + 12

so, x*y = x*[(-12/5)x + 12] = (-12/5)x^2 + 12x = Area, this is the general equation for the integral for finding the area of a rectangle inside your triangle.  Since its completely generic in that it can find the area of any square inside your triangle for any given x, taking the derivitive and setting euqal to zero, will find the max value for the area of a rectangle within the triangle.  The derivitive is:

(-24/5)x + 12 = 0

x = 15/6, and using our equation for y we get,

y = (-12/5)x + 12 = 6 for x = 15/6

Therefore the maximum area is just x*y for these values,

Max Area = (15/6)*6 = 15 cm^2