Question

A cardboard box without a lid is to have a volume of 16,384 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)

Answer 1

Answer:

The dimensions that minimize the amount of cardboard used is

x = 31 cm ,  y = 34 cm & Z = 15.54 cm

Step-by-step explanation:

Volume of the cardboard = 16,384 $cm^{3}$

The function that represents the area of the cardboard without a lid is given by

$f (x,y,z) = xy + 2xz + 2yz$ ------  (1)

Volume of the cardboard with sides x, y & z is

$xyz = 16384$

$z = \frac{16384}{xy}$

Put this value of z in equation (1) we get

$f (x,y,z) = xy + 2x(\frac{16384}{xy} ) + 2y(\frac{16384}{xy} )$

$f (x,y,z) = xy + \frac{32768}{y} + 2y(\frac{32768}{x} )$

Differentiate above equation with respect to x & y we get

$f_{x} = y - \frac{32768}{x^{2} }$

$f_{y} = x - \frac{32768}{y^{2} }$

Take $f_{x} = 0 \ and \ f_{y} = 0$

$y - \frac{32768}{x^{2} } = 0$

$y = 32768 \ x^{-2}$  ------ (2)

$x - \frac{32768}{y^{2} } = 0$

$x = 32768 \ y^{-2}$  ------- (3)

By solving equation (2) & (3) we get

$x^{3} = 32768$

x = 31 cm

From equation 2

$y = 32768 \ x^{-2}$

y = 32768 ($31^{-2}$)

y = 34 cm

$z = \frac{16384}{xy}$

$z = \frac{16384}{(34)(31)}$

Z = 15.54 cm

Thus the dimensions that minimize the amount of cardboard used is

x = 31 cm ,  y = 34 cm & Z = 15.54 cm