1. ## Applied Optimization

I am not getting the correct answer for the following problem, my professor indicated that we should be setting up these problems with a diagram, constraint equation(fixed constraint) and an objective equation, and then solve for 1 variable in the constraint equation, he used Y, and then substitute that into the objective equation, then find the derivative, etc. Here is the problem:

Construction Cost: A carpenter has been asked to build an open box with a square base. The sides of the box will cost $3 per square meter, and the base will cost$4 per square meter. What are the dimensions of the box of greatest that can be constructed for $48? The answer is supposed to be: 2 by 2 by 4/3 meters 2. Originally Posted by kdogg121 I am not getting the correct answer for the following problem, my professor indicated that we should be setting up these problems with a diagram, constraint equation(fixed constraint) and an objective equation, and then solve for 1 variable in the constraint equation, he used Y, and then substitute that into the objective equation, then find the derivative, etc. Here is the problem: Construction Cost: A carpenter has been asked to build an open box with a square base. The sides of the box will cost$3 per square meter, and the base will cost $4 per square meter. What are the dimensions of the box of greatest that can be constructed for$48?

The answer is supposed to be: 2 by 2 by 4/3 meters
Let the side of the base be $b$, and the height be $h$.

Then the volume is:

$V(b,h)=b^2h$

and the cost is:

$C(b,h)=4\times b^2+4 \times 3 \times b\times h$

The cost constaint give:

$4\times b^2+4 \times 3 \times b\times h=48$

Now find $h$ as a function of $b$ from this constaint equation and substitute that
into the objective (the volume $V(b,h)$ ) to get the objective in terms of only
the side of the base $b$.

Now you have a function of only one variable to find the maximum of. You
do this in the usual manner of differentiating and setting the derivative to
zero and solving that equation for the variable, then back substituting to get
the corresponding value/s for the objective, and the constraint to get the
corresponding value of the other variable if you want that as well.

RonL

3. Hello, kdogg121!

A carpenter has been asked to build an open box with a square base.
The sides of the box will cost $3/mē, and the base will cost$4/mē.
What are the dimensions of the box of greatest volume
that can be constructed for $48? The answer is supposed to be: 2 by 2 by 4/3 meters Code:  *-------* /| /| / | / | *-------* |y | | | | | | y| | * | | /x | |/ *-------* x The base has an area of $x^2$ mē. . . At$4/mē, its cost is: . $4x^2$ dollars.

The four sides have an area of $4xy$ mē.
. . At $3/mē, their cost is: . $12xy$ dollars. The total cost is: . $4x^2 + 12xy$ which is limited to$48.
. . There is our constraint: . $4x^2 + 12xy \:=\:48\quad\Rightarrow\quad y \:=\:\frac{48 - 4x^2}{12x}$ .[1]

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

Substitute [1] into [2]: . $V \;=\;x^2\left(\frac{48-4x^2}{12x}\right)$
. . And we have: . $V \;=\;\frac{1}{3}(12x - x^3)$

And that is the function we must maximize . . .

4. So at this point, I would need to take the derivative of the Volume function? What else would I do to get the dimensions? Thanks...

5. Originally Posted by CaptainBlack
.. Now you have a function of only one variable to find the maximum of. You
do this in the usual manner of differentiating and setting the derivative to
zero and solving that equation for the variable, then back substituting to get
the corresponding value/s for the objective, and the constraint to get the
corresponding value of the other variable if you want that as well.
Originally Posted by kdogg121
So at this point, I would need to take the derivative of the Volume function? What else would I do to get the dimensions? Thanks...
You have already been told what to do next

RonL