# Optimization - rectangle inscribed in a right triangle

1. Dec 18, 2013

### ghostanime2001

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
1. $x^{2}+y^{2}=w^{2}$ in terms of $w=\sqrt{x^{2}+y^{2}}$

2. $\dfrac{x}{3}=\dfrac{y}{4}$

3. $\dfrac{l}{3}=\dfrac{4-y}{5}$

4. $\dfrac{l}{x}=\dfrac{4-y}{w}$

3. The attempt at a solution

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

rewriting equation 2 for $x$ we can get an expression in $y$

$4x=3y$

$x=\dfrac{3y}{4}$

which we can substitute in equation 1 to find width $w$ in terms of $y$

$w=\sqrt{x^{2}+y^{2}}$

$w=\sqrt{\left(\dfrac{3y}{4}\right)^{2}+y^{2}}$

$w=\sqrt{\dfrac{9}{16}y^{2}+y^{2}}$

$w=\sqrt{\dfrac{25}{16}y^{2}}$

$w=\dfrac{5}{4}y$

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 $y$ using equation 3.

$\dfrac{l}{3}=\dfrac{4-y}{5}$

$5l=3\left(4-y\right)$

$l=\dfrac{3\left(4-y\right)}{5}$

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

$A\left(y\right)=l\left(y\right)w\left(y\right)$

$A\left(y\right)=\dfrac{3\left(4-y\right)}{5}\dfrac{5}{4}y$

$A\left(y\right)=\dfrac{15y\left(4-y\right)}{20}$

$A\left(y\right)=\dfrac{3y\left(4-y\right)}{4}$

$A\left(y\right)=\dfrac{12y-3y^{2}}{4}$

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

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

solving for $y$ gives

$\dfrac{3}{2}y=3$

$y=2$

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

$l=\dfrac{3\left(4-\left(2\right)\right)}{5}=\dfrac{6}{5}$

$w=\dfrac{5}{4}\left(2\right)=\dfrac{5}{2}$

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

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

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

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2. Dec 18, 2013

### scurty

That works! I get an answer of 3. Did you try it out and see how the answers compared? You can use the proportion $\frac{x}{3} = \frac{y}{4}$ to get the equation in terms of only x or y and then take the derivative, etc.

As for more "correct," every method that arrives at a solution is correct, but this way might be more elegant because it only takes 5 lines or so of equations.

3. Dec 18, 2013

### Ray Vickson

It is a lot easier to use
$$\frac{l}{4-y} = \frac{3}{5} \Longrightarrow l = (3/5)(4-y)\\ \frac{w}{y} = \frac{5}{4} \Longrightarrow w = (5/4)y$$
giving $A = l w = (3/4) y(4-y) = 3 y - (3/4)y^2.$

4. Dec 19, 2013

### ghostanime2001

Ray Vickson, I had overlooked this relation! Your relation directly solves $w$ in terms of $y$ avoiding going through using pythagorean theorum and equation 2. Am I right ? or am I right ?

The first relation is the same one I acquired. It is just written differently.

$\dfrac{w}{y}=\dfrac{5}{4}\Rightarrow w=\dfrac{5}{4}y$

scurty, I have used equation 4 and equation 2 to solve for area $lw$ in terms of $x$ (instead of $y$ as previously used) and I get

$A=lw=4x-\dfrac{4}{3}x^{2}$

compared to $A=lw=3y-\dfrac{3}{4}y^{2}$

5. Dec 19, 2013

### scurty

I can't tell if you are just acknowledging that the method worked (if so, good job!) or you are confused. Sorry, haha, could you just clear that up for me?

6. Dec 19, 2013

### ghostanime2001

the area equation in terms of $x$ is correct, no ? I just want to know if I did my algebra right. I substituted the value of $x=\dfrac{3}{2}=\text{1.5 cm}$ into the area equation and it comes out to $\text{3 }cm^{2}$. Do you have the same numbers & answer ?

7. Dec 19, 2013

### scurty

Yes, it's correct! There's more than one way to arrive at the solution.