Optimization: Find area of largest rect. inscribed inside...

[MarkSA]

We just took our calc I finals and I had some trouble with an optimization problem. There is probably a very easy way to do it but I just couldn't see it. The problem is similar to this:

Find the largest area of a rectangle inscribed in a right triangle with side lengths 3cm and 4cm. Two of the rectangle's sides lie along the triangle's legs.

I drew the picture fine, and labeled one side of the rectangle 3-x and another side 4-y. This made up a portion of the triangle's leg. I used x/y for the other portion of the triangle's leg.

I know the general procedure for optimization - find an equation for what is being optimized, the area of the rectangle in this case, and get it into terms of only one variable. Take the derivative, find where the derivative equals zero, and verify it is a max or min (max in this problem.) But I just couldn't figure out how to set the equation up.

 

[soroban]

Re: Optimization problem

Hello, MarkSA!

I believe you labeled the diagram incorrectly . . .

Find the largest area of a rectangle inscribed in a right triangle with side lengths 3cm and 4cm.
Two of the rectangle's sides lie along the triangle's legs.

        |
      3 *
        |   *  (x,y)
        * - - - *
        |       |   *
        |       |y      *
        |       |           * 
    - - | - - - * - - - - - - - * - - 
        |   x                   4
        |

\(\displaystyle \text{The equation of the line through }(4,0)\text{ and }(0,3)\text{ is: }\;y \;=\;-\frac{3}{4}x + 3\;\;{\bf[1]}\)


\(\displaystyle \text{The area of the rectangle is: }\;A \;=\;xy\;\;{\bf[2]}\)


\(\displaystyle \text{Substitute }{\bf[1]}\text{ into }{\bf[2]}\text{: }\;A \;=\;x\left(-\frac{3}{4}x+3\right)\quad\Rightarrow\quad A \;=\;-\frac{3}{4}x^2 + 3x\)

\(\displaystyle \text{Differentiate and equate to 0: }\;A' \;=\;-\frac{3}{2}x + 3 \:=\:0\quad\Rightarrow\quad x \:=\:2\)

\(\displaystyle \text{Substitute into }{\bf[1]}\text{: }\;y \:=\:-\frac{3}{4}(2) + 3\quad\Rightarrow\quad y \:=\:\frac{3}{2}\)


\(\displaystyle \text{Therefore, the maximum area is: }\;A \;=\;2\cdot\frac{3}{2} \;=\;3\)

 

[TchrWill]

Re: Optimization problem

+y

A
*
*...*
*.......*
*.......*...*
*.......*.......*
********************C.....+x
B

BC = 4 and BA = 3

A = x(4-x)3/4 = 3x - (3/4)x^2

dA/dx = 3 - 1.5x = 0 making x = 2 and y = 1.5





A = x(4 - x)3/4 = [12x - 3x^2]/4 = 3x - (3/4)x^2

dA/dx = 3 - 1.5x = 0 making x = 2

 

[MarkSA]

Re: Optimization problem

Thanks for the reply. It never occurred to me to plot them on a coordinate axis and try to find equation of the line.

I somehow did end up with the right answer for x at the end, but not the right answer for y. I had x = 2, y = 4/3

Well, hopefully they will be forgiving with the partial credit. It's such a bad feeling when you know how to do these general types of problems but can't manage to solve the one on the exam/final, and it's worth 9% of the grade.