Question

How do you find the rectangle of maximum area that can be inscribed in a right triangle with legs of length 3 and 4 if the sides of the rectangle are parallel to the legs of the triangle?

Answer 1

We write the area as a function of the width of the rectangle, which turns out to be quadratic which we maximize by completing the square, giving width `2,` height `3/2,` area `3.`

Explanation:

Let's pin our right triangle on the Cartesian plane with vertices `A(0,0), B(4,0), C(0,3)` so the right angle is at the origin.

We'll place one corner of the rectangle at the origin as well, and sit the rectangle on the x axis so we have another corner at `(x,0).`

The hypotenuse is a line through `(4,0)` and `(0,3)` so is

`(y - 0)(0-4) = (x-4)(3-0) quad or quad y= -3/4 x + 3`

So we have a rectangle with width `x` and height `y` so an area

`A = xy = x (-3/4 x + 3) = -3/4 x^2 + 3x`

We could get the maximum with calculus, but this is asked as an algebra question so we complete the square. First we take out the factor on `x^2` which students sometimes find the trickiest part.

`A = -3/4( x^2 - 4 x)`

Then we halve the coefficient on `x`, and square it, making that the constant. We have to subtract it on the outside with the factor to keep our equation true.

`A = -3/4( x^2 - 4x + 4) - (-3/4) (4)`

`A = -3/4 (x-2)^2 + 3`

We see `A` is maximized when `x=2` giving a maximal `A=3.`