# Optimizing a rectangle

• Feb 18th 2009, 04:22 PM
calc_help123
Optimizing a rectangle
Hi, Im having a problem finishing off a problem. It is a calc ab problem

A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and 5$ for the sides, what is the cost of the least expensive tank?

So the information i gathered was:
SA=LW+2(LH+WH)
W=4
V=36

S0 since V=LWH=36, 4LH=36, L=9/H
Then, SA= 36/h+18+8h
At first i thought I would just need to optimize that and go -36h^-2 + 8 and found the min was at 2.121 but then I thought you might need to factor in the price using another equation. Could someone help me with this problem?
• Feb 18th 2009, 05:30 PM
Soroban
Hello, calc_help123!

Quote:

A tank with a rectangular base and rectangular sides is to be open at the top.
It is to be constructed so that its width is 4 m and its volume is 36 m³.
If building the tank costs $10/m² for the base and$5/m² for the sides,
what is the cost of the least expensive tank?

We can not separate the surface area and the cost of materials.
. . We must come up with a single Cost Function ... which we will minimize.
Code:

* - - - - *
/|        /|
/ |      / | y
* - - - - *  |
|        |  *
y |        | /
|        |/ 4
* - - - - *
x

The width of the box is 4 m.
Let $x$ = length.
Let $y$ = height.

The volume will be 36 m³.
So we have: . $4xy \:=\:36 \quad\Rightarrow\quad y \:=\:\frac{9}{x}$ .[1]

Now we will create a Cost Function.

The front and back has area: . $2 \times xy \:=\:2xy$ m².

The left and right has area: . $2 \times (4y) \:=\:8y$ m².

The total area of the sides of the box is: . $2xy + 8y$ m².

. . At $5/m², the sides will cost: . $5(2xy + 8y) \:=\:10xy + 40y$ dollars. The base has area: . $4x$ m². . . At$10/m², its cost is: . $10(4x) \:=\:40x$ dollars.

Hence, the total cost is: . $C \;=\;10xy + 40y + 40x$ .[2]

Substitute [1] into [2]: . $C \;=\;10x\left(\frac{9}{x}\right) + 40\left(\frac{9}{x}\right) + 40x$

. . and we have: . $C \;=\;90 + 360x^{-1} + 40x$

And that is the function we must minimize . . .