from a cardboard box 12 in by 8 inches are cut out so the sides can be folded up to make a box . What dimentions will yield a maximum volume? What is maximum volume?

we got our V = 4x^3--40x^2+96x
V'=12x^2-80x +96
then we know to use quadratic equation
I get 20( +/-) sqrt 112/6
this is where stuck please help thanks

you have two answers. One will be greater than 8, so it is useless. So, use

(10-2√7)/3 = 1.57
max v = 64(10+7√7)/27

the quadratic solutions are relative max/min points on the graph (you obviously want the max)

test the solutions in the original equation

x < 4 because of the 8 in side

To find the dimensions that yield a maximum volume, we need to find the value of x that maximizes the volume function V(x) = 4x^3 - 40x^2 + 96x.

First, find the derivative of V(x) by taking its derivative with respect to x:
V'(x) = 12x^2 - 80x + 96.

To determine the maximum or minimum points, we need to find the critical points. These are the values of x where the derivative is equal to zero or does not exist.

Setting V'(x) = 0, we have:
12x^2 - 80x + 96 = 0.

Now we can solve this quadratic equation to find the critical points. The quadratic formula is:
x = (-b ± sqrt(b^2 - 4ac)) / 2a.

In this case, a = 12, b = -80, and c = 96, so we can substitute these values into the quadratic formula and simplify:

x = (-(-80) ± sqrt((-80)^2 - 4(12)(96))) / (2(12))
= (80 ± sqrt(6400 - 4608)) / 24
= (80 ± sqrt(1792)) / 24
= (80 ± sqrt(7 * 256)) / 24
= (80 ± 16sqrt(7)) / 24
= (5 ± sqrt(7)) / 3.

Therefore, the critical points are x = (5 - sqrt(7)) / 3 and x = (5 + sqrt(7)) / 3.

Now, to determine whether these critical points correspond to a maximum or minimum, we can use the second derivative test. If the second derivative is positive at a critical point, it corresponds to a minimum. If the second derivative is negative, it corresponds to a maximum.

To find the second derivative, differentiate V'(x):
V''(x) = 24x - 80.

Now we can substitute the critical points into V''(x) to find their corresponding values:

V''((5 - sqrt(7)) / 3) = 24((5 - sqrt(7)) / 3) - 80
= (120 - 24sqrt(7) - 240) / 3
= (120 - 240 - 24sqrt(7)) / 3
= (-120 - 24sqrt(7)) / 3
= -40 - 8sqrt(7).

V''((5 + sqrt(7)) / 3) = 24((5 + sqrt(7)) / 3) - 80
= (120 + 24sqrt(7) - 240) / 3
= (120 - 240 + 24sqrt(7)) / 3
= (-120 + 24sqrt(7)) / 3
= -40 + 8sqrt(7).

Since the second derivative at (5 - sqrt(7)) / 3 is negative (-40 - 8sqrt(7)) and the second derivative at (5 + sqrt(7)) / 3 is positive (-40 + 8sqrt(7)), this tells us that (5 + sqrt(7)) / 3 corresponds to a maximum and (5 - sqrt(7)) / 3 corresponds to a minimum.

Therefore, the dimension that yields a maximum volume is x = (5 + sqrt(7)) / 3. To find the corresponding dimensions of the box, we can substitute this value of x back into the equation for V(x):

V(x) = 4x^3 - 40x^2 + 96x.

V((5 + sqrt(7)) / 3) = 4((5 + sqrt(7)) / 3)^3 - 40((5 + sqrt(7)) / 3)^2 + 96((5 + sqrt(7)) / 3).

Simplifying this expression will give you the maximum volume.