# A box, open at the top is to be made from cardboard. The base of the box is a square of side x and its height is y. If the volume of the box is 32u^2, find the dimensions of the box if the area is to be least. Please help?

May 29, 2018

#### Explanation:

The indicated volume does not make any sense and needs to be
clarified, but for now we can assume the volume to be $32 {m}^{3}$:

The area of the base is:

${A}_{b a s e} = {x}^{2}$

The volume would be:

$V = {x}^{2} y = 32 {m}^{3}$

Calculating $y$ in terms of $x$ in above equation results:

$y = \frac{32}{x} ^ 2$

Total surface area of a box with open top is:

$A = {x}^{2} + 4 x y$

Substituting for $y$:

$A = {x}^{2} + 4 x \cdot \frac{32}{x} ^ 2$

$A = {x}^{2} + \frac{128}{x}$

To find the critical points equate the first derivative to zero:

$\frac{\mathrm{dA}}{\mathrm{dx}} = A ' = 2 x - \frac{128}{x} ^ 2$

$\frac{2 {x}^{3} - 128}{x} ^ 2 = 0$

${x}^{3} - 64 = 0$

${x}^{3} = 64$

$x = 4 m$

To verify the nature of the critical point use 2nd derivative test:

A"=2+256/x^3

A"(4)=2+256/4^3=6>0=> Verifies it is a minimum so:

$x = 4 m$

$y = \frac{32}{16} = 2 m$

Thus the box dimensions for a minimum surface area are:

$4 m \cdot 4 m \cdot 2 m$

And the minimum surface area would be:

$A = {x}^{2} + 4 x y = {4}^{2} + 4 \cdot 4 \cdot 2 = 16 + 32 = 48 {m}^{2}$

May 30, 2018

x=4root3[u^2............ y=2root3[u^2

#### Explanation:

Assuming the units expressed for the volume are correct and are indeed expressed in terms of ${u}^{2}$ then we may say ,

Volume $V = {x}^{2} y = 32 {u}^{2}$........$\left[1\right]$ , the surface area $A = {x}^{2} + 4 x y$......$\left[2\right]$.

From ....$\left[1\right]$ $y = 32 {u}^{2} / {x}^{2}$ and substituting this value for $y$ in .....$\left[2\right]$,

$A = {x}^{2} + 4 x \frac{32 {u}^{2}}{{x}^{2}}$ = [x^2+[4][32u^2]]/x]......$\left[3\right]$

Differentiating ......$\left[3\right]$ wrt $x$, $\frac{d}{\mathrm{dx}}$= $2 x - 4 \frac{32 {u}^{2}}{x} ^ 2$ =0 [ for max/min and considering $u$ constant]

${x}^{3} = 64 {u}^{2}$, therefore x = 4root3[u^2. substituting this value for $x$ in .....$\left[1\right]$ will give the value of $y$ in the answer.

The second derivative =$2 + 8 \frac{32 {u}^{2}}{x} ^ 3$ which is positive, since $x$ and $u$ must be both positive [ must have positive values for a box] and so confirms that the value of $x$ obtained will minimise the surface area of the box.