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?

1 Answer
Apr 27, 2018

Answer:

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.#