Question

Use Lagrange multipliers to find the dimensions of a rectangular box with largest volume if the total surface area is given as 400 cm2. (Enter the dimensions (in centimeters) as a comma separated list.)

Answer 1

Answer:

The dimension of the given box is $\sqrt{\frac{200}{3}}\ cm$ by $\sqrt{\frac{200}{3}}\ cm$ by$\sqrt{\frac{200}{3}}\ cm$.

Step-by-step explanation:

Assume the dimension of the box is x by y by z.

Total surface area of the box= 2xy+2yz+2xz=400

The volume of the box is f(x,y,z)= xyz

To maximum volume under constrain,

g(x,y,z)=2xy+2yz+2xz-400=0

we use Langrange Multiplier.

Assume that,

$\triangledown f=\lambda \triangledown g$

$\Rightarrow =\lambda$

$\Rightarrow =\lambda$

$\therefore xy=2 \lambda (y+z)$ $\Rightarrow xyz= 2\lambda(xy+xz)$

$\therefore yz=2 \lambda (x+z)$  $\Rightarrow xyz= 2\lambda (xy+yz)$

$\therefore xy=2 \lambda (x+y)$  $\Rightarrow xyz= 2\lambda (xz+yz)$

$\therefore xyz= 2\lambda (xy+xz)=2 \lambda (xy+yz)=2 \lambda (xz+yz)$

If $\lambda \neq 0$,

$\Rightarrow (xy+xz)= (xy+yz)=(xz+yz)$

This implies that, x=y=z

Then 2xy+2yz+2xz=400

⇒xy+yz+xz=200

⇒x²+x²+x²=200

⇒3x²=200

$\Rightarrow x=\sqrt {\frac{200}{3}}$

Therefore $x=y=z=\sqrt{\frac{200}{3}}$ cm.