Help with Applied Optimization Problems

[roodypoo78]

Well, it seems that I've become stuck with a couple of these wonderful homework problems, and I sure could use some helpful advice/hints/solutions!

First:

A rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be a maximum?


Next:

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

I've spent the last 4 hours trying to figure out the answers, and I keep getting answers that don't work, or don't make sense. :evil: On the first question, I keep getting dimensions of 200x200, which is obviously wrong, and on the second question, I can't even seem to set up a correct equation because the derivative I get is way messed up.

Any help is greatly appreciated, and might even garner you a giant THANK YOU.

Thanks.

 

[soroban]

He;;o, roodypoo78!

A rancher has 400 feet of fencing with which to enclose two adjacent rectangular corrals.
What dimensions should be used so that the enclosed area will be a maximum?

      * - - - - * - - - - *
      |         |         |           Let x = length
     y|        y|         |y
      |         |         |           Let y = width
      * - - - - * - - - - *
      :.........x.........:

There are 2 sections which are \(\displaystyle x\) feet long and 3 sections which are \(\displaystyle y\) feet long.

. . Hence, the total fencing is: .\(\displaystyle 2x\,+\,3y\:=\:400\;\;\Rightarrow\;\;y\:=\:\frac{400\,-\,2x}{3}\) .[1]

The total area is: .\(\displaystyle A\;=\;xy\) .[2]

. . Substitute [1] into [2]: .\(\displaystyle A\;=\;x\left(\frac{400\,-\,2x}{3}\right)\;\;\Rightarrow\;\;A\;= \;\frac{1}{3}(400x\,-\,2x^2)\)

And that is the function we must maximize.

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches.
What dimensions will produce a box with maximum volume?

              * - - - - - - - - *
            / |               / |
          /   |             /   | y
        /     |           /     |        Let x = length and width
      * - - - - - - - - *       *
      |                 |     /          Let y = height of the box
    y |                 |   / x
      |                 | / 
      * - - - - - - - - *
                x

The area of the base is: \(\displaystyle x^2.\)
The four sides have an area of: \(\displaystyle 4xy\)
. . The total area is: .\(\displaystyle x^2\,+\,4xy\:=\:108\;\;\Rightarrow\;\;y\,=\;\frac{108\,-\,x^2}{4x}\) .[1]

The volume of the box is: .\(\displaystyle V\;=\;x^2y\) .[2]

Substitute [1] into [2]: .\(\displaystyle V\;= \;x^2\left(\frac{108\,-\,x^2}{4x}\right)\;=\;\frac{1}{4}(108x\,-\,x^3)\)

And that is the function we must maximize . . .

 

[roodypoo78]

Thank You very much.

Thank you so much for your help. Saved me hours and tons of Tylenol!

-Gavin