Question

A cardboard box without a lid is to have a volume of 10,976 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:

28cm*28cm*14cm

Step-by-step explanation:

Five side box for cardboard are: xy, 2xz and 2yz

so the function of Area will be:

$f(x,y.z)=xy+2xz+2yz$   equation 1

The Volume will be:

$xyz=10976 cm^{3}$        equation 2

we can make z the subject of the formula by divide both side by xy

now   z = $\frac{10976}{xy}$       equation 3

To eliminate z, substitute z into equation 1

f(x,y)=xy+2x$(\frac{10976}{xy})$+2y

f(x,y)=xy+$\frac{21952}{y}$+$\frac{21952}{x}$

Now we have to derivative of x and y

$f_{x}= y-\frac{21952}{x^{2} }$

$f_{y}=x-\frac{21952}{y^{2} }$

for y and x

$y-\frac{21952}{x^{2} } =0\\y=21952x^{-2} \\x-\frac{21952}{y^{2} } =0\\x=21952y^{-2}$

substitute y into x

$x=21952(21952x^{-2} )^{-2} \\x=(21952)^{-1} x^{4} \\ x^{3} =21952\\cube-root-both side\\x=\sqrt[3]{21952} \\x=28$

to find y

$y=21952(28)^{-2} \\y=28$

to find z

$z=\frac{10976}{xy}$

z=$\frac{10976}{28*28}$

$z=14$

so The dimensions are 28cm*28cm*14cm