1. ## World of Functions

A open box with a square base is required to have a volume of 10 cubic feet.

(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.

(b) How much material is required for a base 1 foot by 1 foot?

2. Originally Posted by magentarita
A open box with a square base is required to have a volume of 10 cubic feet.

(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.

(b) How much material is required for a base 1 foot by 1 foot?
Let h denote the height of the box. Then the volume of the box is:

$V = x^2 \cdot h = 10\ cft~\implies~ \boxed{h = \frac{10}{x^2}}$

The surface area of the box exists of 1 square and 4 congruent rectangles:

$A = x^2+4 \cdot h \cdot x$ Plug in the term for h from the first equation to eliminate h:

$A(x)=x^2+4 \cdot \frac{10}{x^2} \cdot x = x^2+\frac{40}x$

To calculate the extreme amount of material use the first derivation of A:

$A'(x) = 2x-\frac{40}{x^2}$ . Solve the equation $A'(x) = 0$ for x. For your confirmation only: $x = \sqrt[3]{20} \approx 2.71\ ft$

Where did x^2 come from?

Also, isn't the volume of a box V = length times width times height?

Thanks,
Rita

4. Originally Posted by magentarita
Where did x^2 come from?

Also, isn't the volume of a box V = length times width times height?

Thanks,
Rita
Your problem said: "A open box with a square base is ..."

That means the length and the width are equal. Since one side of the square should have been labeled by x the area of the base is $x^2$ and the volume of the box is calculated by:

$volume = (base\ area) \cdot (height)$ or using one letter variable names:

$v = x^2 \cdot h$

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6. ## Re: World of Functions

so how do you express the ammount of material that is needed as a function of the length x. if u allready explained please elaborate

7. ## Re: World of Functions

We have:

$x^2h=V$ where $V$ is a given constant. Hence:

$h=\frac{V}{x^2}$

The amount of material $A$ is the area of the bottom and 4 sides:

$A=x^2+4xh$ and so:

$A(x)=x^2+4x\left(\frac{V}{x^2} \right)=x^2+\frac{4V}{x}$