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.)

Respuesta :

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:

[tex]f(x,y.z)=xy+2xz+2yz[/tex]   equation 1

The Volume will be:

[tex]xyz=10976 cm^{3}[/tex]        equation 2

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

now   z = [tex]\frac{10976}{xy}[/tex]       equation 3

To eliminate z, substitute z into equation 1

f(x,y)=xy+2x[tex](\frac{10976}{xy})[/tex]+2y

f(x,y)=xy+[tex]\frac{21952}{y}[/tex]+[tex]\frac{21952}{x}[/tex]

Now we have to derivative of x and y

[tex]f_{x}= y-\frac{21952}{x^{2} }[/tex]

[tex]f_{y}=x-\frac{21952}{y^{2} }[/tex]

for y and x

[tex]y-\frac{21952}{x^{2} } =0\\y=21952x^{-2} \\x-\frac{21952}{y^{2} } =0\\x=21952y^{-2}[/tex]

substitute y into x

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

to find y

[tex]y=21952(28)^{-2} \\y=28[/tex]

to find z

[tex]z=\frac{10976}{xy}[/tex]

z=[tex]\frac{10976}{28*28}[/tex]

[tex]z=14[/tex]

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