Chocolate Box Company is going to make open-topped boxes out of 3 × 12-inch rectangles of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way? (Round your answer to the nearest tenth.)

what i got so far...
4x^3-30x^2+36x

in^3=?

Question

Chocolate Box Company is going to make open-topped boxes out of 3 × 12-inch rectangles of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way? (Round your answer to the nearest tenth.)

what i got so far...
4x^3-30x^2+36x

in^3=?

anonymous · Answer

my answer x=1/2(5-sqr(13)=0.697224 <---

x=1/2(5+sqr(13)=4.30278 

thus in^3= 11.9

anonymous · Answer

i found the solution already for the original problem but can u guys help me with another one.

A postal service will accept packages only if the length plus girth is no more than 60 inches. (See the figure.) Assuming that the front face of the package (as shown in the figure) is square, what is the largest volume package that the postal service will accept?

anonymous · Answer

would it be v"=120x-24x

anonymous · Answer

Volume formula
$$V=width*lenght*height$$
A drawing will help.....REALLY help....
but here's the math

$$width=3-x$$
$$height=x$$
$$length=12-x$$

So,
$$V=(3-x)(x)(12-x)       \;\;\;\;\;\;\;\; substitution$$
$$V=(3x-x^{2})(12-x)$$
$$V=x^{3}-12x^{2}-3x^{2}+36x$$
$$V=x^{3}-15x^{2}+36x$$

Derivative
$$V\prime=3x^{2}-30x+36$$
$$V\prime=3(x^{2}-10x+12)$$

Set equal to 0, and
$$x=\{\sqrt{13}+5, -\sqrt{13}+5\}$$

Since the first solution is bigger then on of the sides of the cardboard that one is not it, so  the second value is what we are looking for , go back and plug in accordingly for height, width and length to get the dimensions. OR since you want the Volume only.....just plug in that value into the Volume formula above for x.

anonymous · Answer

Pardon my grammar am typing on the fly here.