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.