[Calculus]Optimization question?

For optimization, you must use the method of Lagrange multiplier. Here is the method, Let b be the base and h be the height.
Volume= b^2 h.
Surface area = 4 bh+b^2=108. Now define a function
f(b,h) =b^2h-l(4bh+b^2-108)
Take partial derivative wrt b and h. You will have\
wrt b   2bh -l(4h+2b)=0
wrt h, b^2 -4bl=0. This gives b=4l.
Putting this value in the first equation, you get h=2l. Now solve for l
4  4 l 2l+16 l^2=108. This give l= +/- 3/2. Neglect the negative.
Therefore you get b= 4l=6, height = 3.
Since we take only first derivative, the values given can be a maximum or minimum. Therefore, it is called optimisation.

Let the base have sides of b and let the box have a height of h

108 = b^2 + 4 * b * h
V = b^2 * h

Let's relate b to h

108 = b^2 + 4bh
108 - b^2 = 4bh
h = (108 - b^2) / (4b)

V = b^2 * h
V = b^2 * (108 - b^2) / (4b)
V = (1/4) * b * (108 - b^2)
V = (1/4) * (108b - b^3)

Now derive

dV/db = (1/4) * (108 - 3b^2)
dV/db = (3/4) * (36 - b^2)
dV/db = (3/4) * (6 - b) * (6 + b)

Let dV/db = 0.  This tells us when the change in volume related to the change in the base dimensions reaches a critical point.

0 = (3/4) * (6 - b) * (6 + b)
0 = (6 - b) * (6 + b)

6 - b = 0
b = 6

6 + b = 0
b = -6

b > 0, so b = -6 doesn't make any sense

b = 6

108 = b^2 + 4bh
108 = 36 + 24h
72 = 24h
3 = h

V = b^2 * h
V = 6^2 * 3
V = 36 * 3
V = 108

You want a box with a square base and a fixed surface area of 108 in² with an open top.

So we'll create that equation for surface area with one base and 4 sides:

A = s² + 4hs

A is known, h and s are unknown, so:

108 = s² + 4hs

Let's solve this for h in terms of s, since we'll need it later:

108 - s² = 4hs
27/s - s/4 = h

Now we want the volume equation:

V = hs²

We have an expression for h in terms of s, so:

V = (27/s - s/4)s²
V = 27s - s³/4

To find the value of "s" to give a maximum volume, solve for the zero of the first derivative:

dV/ds = 27 - 3s²/4
0 = 27 - 3s²/4
3s²/4 = 27
3s² = 108
s² = 36
s = ±6

We can't have a negative length, so drop the negative to leave:

s = 6 in

Now that we have s, solve for h:

h = 27/s - s/4
h = 27/6 - 6/4
h = 9/2 - 6/4
h = 18/4 - 6/4
h = 12/4
h = 3 in

The dimensions of the box is:

6 in sides of the square base and a height of 3 in.