# 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?

Apr 27, 2018

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 $\frac{3}{2} ,$ area $3.$

#### Explanation:

Let's pin our right triangle on the Cartesian plane with vertices $A \left(0 , 0\right) , B \left(4 , 0\right) , C \left(0 , 3\right)$ 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 $\left(x , 0\right) .$

The hypotenuse is a line through $\left(4 , 0\right)$ and $\left(0 , 3\right)$ so is

$\left(y - 0\right) \left(0 - 4\right) = \left(x - 4\right) \left(3 - 0\right) \quad \mathmr{and} \quad y = - \frac{3}{4} x + 3$

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

$A = x y = x \left(- \frac{3}{4} x + 3\right) = - \frac{3}{4} {x}^{2} + 3 x$

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.

$A = - \frac{3}{4} \left({x}^{2} - 4 x\right)$

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.

$A = - \frac{3}{4} \left({x}^{2} - 4 x + 4\right) - \left(- \frac{3}{4}\right) \left(4\right)$

$A = - \frac{3}{4} {\left(x - 2\right)}^{2} + 3$

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