
cost minimization
I need some help with a problem involving the optimization of area.
This is the prompt to the problem:
"A closed box with square base is to be built to house an ant colony. The bottom of the box and all four sides will be made of material costing $1 per square foot, and the top is to be constructed of glass costing $5 per square foot. What are the dimensions of the box of greatest volume than can be constructed for $72?"
I have worked several optimization problems both involving cost and not but in this problem I am given the cost and not the area. I understand the process of optimization: Construct an equation that models the problem (hopefully with a sketch), use the given value in its respective equation and solve for one of the variables, plug this quantity into the main equation and solve for the other variable.
I guess I should start with checking if my equation is correct.
Surface Area = sides + top + bottom
SA = 4xy + x^2 + x^2 Volume = x^2*y
Cost = 4xy(1) + x^2(5) + x^2(1) Given value : $72 (max cost)
Cost = 4xy + 5x^2 + x^2
With problems like this I usually just take the derivative and doing that gave me the wrong dimensions. So going back I think that I need to set the cost equation equal to 72. However that would then incorporate implicit differentiation into an optimization problem and I have never done that before, just implicit by itself. Am I headed in the right direction and if so then what?

Let x be depth, y be width, and z be height. Then,
$\displaystyle xy+2xz+2yz+5xy=72$
$\displaystyle 2xz+2yz+6xy=72$
You can solve this using partial derivatives, by writing volume as a function of only two variables x, and y, and taking $\displaystyle \nabla f =0$.
Or you can use the AMGM inequality:
$\displaystyle \frac{2xz+2yz+6xy}{3}\ge(2xz\cdot 2yz\cdot 6xy)^{\frac{1}{3}}$
$\displaystyle \frac{72}{3}\ge(2xz\cdot 2yz\cdot 6xy)^{\frac{1}{3}}$
$\displaystyle 24^3\ge 2xz\cdot 2yz\cdot 6xy$
$\displaystyle 24^3\ge 24x^2y^2z^2$
$\displaystyle 24^2\ge x^2y^2z^2$
$\displaystyle 24\ge xyz $
$\displaystyle V \le 24 \implies V_{max}=24\text{ft}^3$
(The dimensions of your box are the set $\displaystyle \{x,y,z\in\mathbb{R}:xyz=24\}$ )

Thank you for your help Greengoblin. Your final answer is correct according to my answer sheet but you created one more variable than is needed Greengoblin. I always learned that when a square base is involved one variable should be used for both of them. that is why my equation for volume is V=x^2*y. Also I have not learned about the AMGM inequality so that is probably not a method I need to know yet. Also I'm not sure as to how I would solve it with partial derivatives.

You're welcome. I misread the part about having a square base as having a rectangular base, so replacing any instances of xy in the above calculation with $\displaystyle x^2$, should give more appropriate expressions as you pointed out. (This is also why I though partial derivatives would be needed..If you haven't heard of them yet, don't worry..their concerned with differentiating functions of several variables, which would have been the case had we had three independent dimensions.)
The calculus way is:
$\displaystyle x^2+4xy+5x^2=72$
$\displaystyle 6x^2+4xy=72$
Now you can solve for y,
$\displaystyle y=\frac{726x^2}{4x}$
...and get volume as a function of x only, by substitution:
$\displaystyle V(x,y)=x^2y\implies V(x)=\frac{72x^26x^4}{4x}=18x\frac{3}{2}x^3$
Now you can find f'(x)...:)

you know I just realized why I wasn't getting the problem right. I was taking the right approach but seeing the way you set up the equation allowed me to see where my mistake was.
I must have confused the surface area equation with the cost equation after incorporating the $1 and $5 values for their respective faces. I took the derivative of this equation thinking the top and bottom quantities need to stay separate:
[IMG]file:///C:/Users/Robert/AppData/Local/Temp/mozscreenshot2.jpg[/IMG]http://www.mathhelpforum.com/mathhe...73682ef21.gif
Instead of adding them like you did and taking the derivative of this equation:
http://www.mathhelpforum.com/mathhe...60b4fe271.gif
That's what threw me off all along. I considered doing that earlier apparently didn't follow through. Wow definitely one of those "oh......duh" moments. Thank you again I appreciate the help.
