IMathAS-libretexts

An open top box with height

\displaystyle {y}

and a square base with side length

\displaystyle {x}

has a volume of

\displaystyle {62.5}

\displaystyle {c}{m}^{{3}}

. Our goal is to find the dimensions

\displaystyle {x}

and

\displaystyle {y}

that minimize the amount of material used.

(1st) Find a formula for the surface area

\displaystyle {A}

of the box in terms of only

\displaystyle {x}

, the length of one side of the square base.
(Hint: Use the volume formula that relates

\displaystyle {x}

\displaystyle {y}

, and

\displaystyle {62.5}

to isolate the height

\displaystyle {y}

of the box in terms of

\displaystyle {x}

. Substitute the expression for

\displaystyle {y}

into your surface area formula.)

\displaystyle {A}{\left({x}\right)}=

(2nd) Find the derivative,

\displaystyle {A}'{\left({x}\right)}

\displaystyle {A}'{\left({x}\right)}=

(3rd) Calculate the value of

\displaystyle {x}

that makes the derivative zero.
(Hint: Multiply both sides by

\displaystyle {x}^{{2}}

\displaystyle {A}'{\left({x}\right)}={0}

when

\displaystyle {x}=

(4th) We next have to make sure that this value of

\displaystyle {x}

gives a minimum value for the surface area. Let's use the second derivative test.
(a) Find A"

\displaystyle {\left({x}\right)}

.
A"

\displaystyle {\left({x}\right)}=

(b) Evaluate A"

\displaystyle {\left({x}\right)}

at the

\displaystyle {x}

-value you found in the third step above.

Note: Since the second derivative is positive, the graph of

\displaystyle {A}{\left({x}\right)}

is concave up around that value. So the zero of

\displaystyle {A}'{\left({x}\right)}

is a local minimum for

\displaystyle {A}{\left({x}\right)}

.

(5th) What is the height

\displaystyle {y}

of the box?

\displaystyle {y}=

(6th) What is the minimum surface area

\displaystyle {A}

\displaystyle {A}=