Thread: 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?

Hello, calc_help123!

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 = length.
Let = height.

The volume will be 36 m³.
So we have: . .[1]


Now we will create a Cost Function.


The front and back has area: . m².

The left and right has area: . m².

The total area of the sides of the box is: . m².

. . At $5/m², the sides will cost: . dollars.


The base has area: . m².

. . At $10/m², its cost is: . dollars.


Hence, the total cost is: . .[2]


Substitute [1] into [2]: .

. . and we have: .


And that is the function we must minimize . . .