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?

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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?

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Answer 1 · 2018-04-27T10:53:31

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

Explanation:

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

The hypotenuse is a line through $(4, 0)$ $(4,0)$ and $(0, 3)$ $(0,3)$ 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 $A=3.$

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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

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