Question

A box of volume 324 m3 with a square bottom and no top is made of two different materials. The cost of the bottom is $40/m2 and the cost of the sides is $30/m2. Find the dimensions of the box that minimize the total cost.

Answer 1

The dimension of the box is:

Bottom length $3\sqrt[3]{18}$ m

Height $2\sqrt[3]{18}$ m

Differential

If the length of the bottom is x and the height is y. The volume of the box equation will be:

$V=x^{2} y=324$

To find the cost for making the box is measure the surface area of box (1 bottom and 4 side)

$A=x^{2} +4xy$

And if the material of bottom is $40 per meter square and the sides is $30 per meter square, the cost for making the box is:

$cost=40(A_{bottom})+30(A_{sides})\\cost=40(x^{2} )+30(4xy)\\cost=40x^{2} +120xy$

From volume equation we can substitute one variable to make one variable function.

$V=x^{2} y=324$

$x^{2} y=324\\y=\frac{324}{x^{2} }$

Let's substitute y to cost function.

$cost=40x^{2} +120xy\\cost=40x^{2} +120x(\frac{324}{x^{2} } )\\cost=40x^{2} +\frac{38880}{x}$

I know there is big number, so let's make it become exponent.

$cost=40x^{2} +\frac{38880}{x}\\cost=2^{3}5x^{2} +2^{5} 3^{5}5x^{-1}$

To make the minimal cost, we can differentiate the cost function.

$cost=2^{3}5x^{2} +2^{5} 3^{5}5x^{-1}\\\frac{dy}{dx}=(2)2^{3}5x+(-1)2^{5} 3^{5}5x^{-2}\\\frac{dy}{dx}=2^{4}5x-2^{5}3^{5}5x^{-2}$

After we differentiete it, when we find the minimum/ maximum point, we equal the equation to zero.

$2^{4}5x-2^{5}3^{5}5x^{-2}=0\\2^{4}5x=2^{5}3^{5}5x^{-2}\\x^{3}=2.3^{5}\\x=(2.3^{5})^{\frac{1}{3} } \\x=2^{\frac{1}{3} } 3^{\frac{5}{3} }\\x=3\sqrt[3]{2.3^{2} } \\x=3\sqrt[3]{18}$

Then we got the length of the bottom. After that we find the height of the box (y)

$y=\frac{324}{x^{2} }\\y=\frac{324}{(2^{\frac{1}{3}}3^{\frac{5}{3} } )^{2} }\\y=\frac{2^{2} 3^{4} }{2^{\frac{2}{3}}3^{\frac{10}{3} } }\\y=2^{\frac{4}{3} }3^{\frac{2}{3} }\\y=2\sqrt[3]{2.3^{2} } \\y=2\sqrt[3]{18}$

Learn more about differential here: https://brainly.com/question/24062595

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