Question

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.

Answer 1

Explanation:

Let x, be the length of the square box and S be the surface of the box,

Let y, be the height of the square box and V for the volume.

               $V = x^2*y$

The top of the box is open and the material of the box is equal to the surface area of box.

                $S= x^2 + 2xy + 2xy$

then,        $S=x^2+4xy$

Putting the value of S (surface area).

                $10800=x^{2} +4xy$

                $y=\frac{10800 - x^2}{4x}$

subtitutting the following equation in volume at the place of y.

               $V=x^{2}(\frac{10800-x^2}{4x} )$

Then,      $V=2700x*\frac{1}{4}x^3$

Differentiate the volume to 0.

               $V'=2700-\frac{3}{4}x^2$

Then,       $\frac{3}{4}x^2 = 2700$

∴              $x^2 = 3600$

and          $x=60$ $, y=30$

Then, we have to find maximum volume by putting the value of x.

                $V=(2700*60) - (\frac{1}{4}(60) )$

                $V=10,8000 cm^3$

Answer 2

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.