- #1

ghostanime2001

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## Homework Statement

A rectangle is to be inscribed in a right triangle having sides 3 cm, 4 cm and 5 cm, as shown on the diagram. Find the dimensions of the rectangle with greatest possible area.

## Homework Equations

1. [itex]x^{2}+y^{2}=w^{2}[/itex] in terms of [itex]w=\sqrt{x^{2}+y^{2}}[/itex]

2. [itex]\dfrac{x}{3}=\dfrac{y}{4}[/itex]

3. [itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

4. [itex]\dfrac{l}{x}=\dfrac{4-y}{w}[/itex]

## The Attempt at a Solution

Area of the rectangle is [itex]A=lw[/itex]. But, we want an expression for area in terms of one variable.

rewriting equation 2 for [itex]x[/itex] we can get an expression in [itex]y[/itex]

[itex]4x=3y[/itex]

[itex]x=\dfrac{3y}{4}[/itex]

which we can substitute in equation 1 to find width [itex]w[/itex] in terms of [itex]y[/itex]

[itex]w=\sqrt{x^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\left(\dfrac{3y}{4}\right)^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{9}{16}y^{2}+y^{2}}[/itex]

[itex]w=\sqrt{\dfrac{25}{16}y^{2}}[/itex]

[itex]w=\dfrac{5}{4}y[/itex]

Using similar triangles of the upper right, smaller right triangle and the bigger, outer triangle we set a relationship between the base of smaller triangle and the base of the outer triangle, and, the hypotenuse of the smaller triangle and the hypotenuse of the outer triangle. We solve for the length of the rectangle in terms of [itex]y[/itex] using equation 3.

[itex]\dfrac{l}{3}=\dfrac{4-y}{5}[/itex]

[itex]5l=3\left(4-y\right)[/itex]

[itex]l=\dfrac{3\left(4-y\right)}{5}[/itex]

We now have the length [itex]l[/itex] and width [itex]w[/itex] in terms of one variable [itex]y[/itex]. We substitute the expressions in the area of the rectangle to find the area in one variable [itex]y[/itex]

[itex]A\left(y\right)=l\left(y\right)w\left(y\right)[/itex]

[itex]A\left(y\right)=\dfrac{3\left(4-y\right)}{5}\dfrac{5}{4}y[/itex]

[itex]A\left(y\right)=\dfrac{15y\left(4-y\right)}{20}[/itex]

[itex]A\left(y\right)=\dfrac{3y\left(4-y\right)}{4}[/itex]

[itex]A\left(y\right)=\dfrac{12y-3y^{2}}{4}[/itex]

[itex]A\left(y\right)=3y-\dfrac{3}{4}y^{2}[/itex]

Differentiating this expression and setting [itex]A'\left(y\right)=0[/itex] gives [itex]0=3-\dfrac{3}{2}y[/itex]

solving for [itex]y[/itex] gives

[itex]\dfrac{3}{2}y=3[/itex]

[itex]y=2[/itex]

substitute [itex]y=2[/itex] in the length [itex]l=\dfrac{3\left(4-y\right)}{5}[/itex] and width [itex]w=\dfrac{5}{4}y[/itex] equations respectively,

[itex]l=\dfrac{3\left(4-\left(2\right)\right)}{5}=\dfrac{6}{5}[/itex]

[itex]w=\dfrac{5}{4}\left(2\right)=\dfrac{5}{2}[/itex]

Therefore, the rectangle with dimensions [itex]l=\dfrac{6}{5}=\text{1.2 cm}[/itex] and [itex]w=\dfrac{5}{2}=\text{2.5 cm}[/itex] has the maximum possible area of [itex]\text{3 }cm^{2}[/itex]

*above I used the more tedious way I guess...

My question is if I can do it using equation 4 since [itex]l[/itex] and [itex]w[/itex] are already there and cross multiplying would give me area automatically. I would have an expression in [itex]x[/itex] and [itex]y[/itex] but I can use equation 2 to find either [itex]x[/itex] or [itex]y[/itex] and set up the area in one variable. Would this way be the more correct method or give the same answer as above? Thanks !