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:

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The width of the box is 4 m.

Let $\displaystyle x$ = length.

Let $\displaystyle y$ = height.

The volume will be 36 m³.

So we have: .$\displaystyle 4xy \:=\:36 \quad\Rightarrow\quad y \:=\:\frac{9}{x}$ .[1]

Now we will create a Cost Function.

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

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

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

. . At $5/m², the sides will cost: .$\displaystyle 5(2xy + 8y) \:=\:10xy + 40y$ dollars.

The base has area: .$\displaystyle 4x$ m².

. . At $10/m², its cost is: .$\displaystyle 10(4x) \:=\:40x$ dollars.

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

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

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

And **that** is the function we must minimize . . .