Question

Find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box.) (x, y, z)

Answer 1

Answer:

The dimensions of the box are 12 cm , 12 cm , 12 cm

Step-by-step explanation:

Let x , y and z be the dimensions of box

Volume of box =xyz=1728

$z=\frac{1728}{xy}$

Surface area of box = $2xy+2yz+2xz=2xy+2y(\frac{1728}{xy})+2x(\frac{1728}{xy})$

Let $f(x,y)=2xy+2(\frac{1728}{x})+2(\frac{1728}{y})$

To get minimal surface area

$\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial x}=0$

$2y-2(\frac{1728}{x^2})=0$

$y=\frac{1728}{x^2}$ ----1

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial y}=0$

$2x-2(\frac{1728}{y^2})=0\\x=\frac{1728}{y^2} \\y^2=\frac{1728}{x}$

Using 1

$(\frac{1728}{x^2} )^2=\frac{1728}{x}$

x=0 and $x^3=1728$

Side can never be 0

So,$x^3=1728$

x=12

$y=\frac{1728}{x^2} \\y=\frac{1728}{12^2}$

y=12

$z=\frac{1728}{xy}\\z=\frac{1728}{(12)(12)}$

z=12

The dimensions of the box are 12 cm , 12 cm , 12 cm