If 10 800 cm2 of material is available to make a box with a square base and an open top find the largest possible volume of the box.

What are you given variables?
Surface area of box: 10800cm²
Volume of box: s²h
Where:
SA=surface area
s = side of square base
h = height of box

Make the height of the box in terms of s
You can write the formula for the surface area of the box in terms of s and h like so:
S. A. = s² + 4sh
Where: S. A. = surface area or 1200 cm², s² = the square base, and 4sh = the four ’walls’ of the box.
10800 = s² + 4sh
10800 - s² = 4sh
(10800 - s²)/(4s) = h

Substitute h (in terms of s) into the formula for volume.
v(s) = s²((10800- s²)/(4s) // Simplify.
v(s) = s(10800 - s²)/4 // Expand.
v(s) = 2700s - (1/4)s^3 // To find the largest possible volume of the box, you find the maximum value of this function.

Take the derivative of the volume function using the Power Theorem.
v’(s) = 2700 - (3/4)s² // Zeroes of the d/dx v function will give you the x values that correspond to local extrema in the v function.
0 = 2700 - (3/4)s² // Solve for zeroes.
-2700= (-3/4)s²
3600= s²
Your two zero values for s are -60 and 60.

Take the second derivative to see if the s values will give you a local maximum or minimum.
v"(s) = -(3/2)s
v"(-60) = -(3/2)(-60)
v"(-60) = 90 // This indicates a local minimum for v at s, not what we are looking for.
v"(60) = -(3/2)(60)
v"(60) = -90 // This indicates a local maximum for v at s, which is what we are looking for, the maximum volume of the box. This makes sense because if you remember what we assigned the variable s to, a side length, side lengths cannot be negative.

Once we have found the value for s, we can substitute it into the function we created for the volume of the box.
v(s) = 2700s- (1/4)s^3
v(s) = 2700(60) - (1/4)(60)^3
v(s) = 162000- (1/4)(216000)
v = 162000- 54000

The largest possible volume of the box is 108000 cubic centimeters.

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