Question

A carpenter has been asked to build an open box with a square base, where an open box means a box without a top. The sides of the box will cost $3 per square meter, and the base will cost $3 per square meter. What are the dimensions of the box of maximal volume that can be constructed for $144 ?
If x is the length of one of the sides of the base and z is the height of the box. Find a function f(x) for the volume of the box.
NOTE: The function should just be in terms of x.

Answer 1

i guess the base cost
\[3x^2\] because the area is
\[x^2\] and if the height is z then each side is
\[3xz\] so four sides gives
\[12xz\] and you know that
\[3x^2+12xz=144\] so you can solve this for z

Answer 2

I think it's more complicated than that satellite

Answer 3

gives
\[z=\frac{144-3x^2}{12x}\] or if you prefer
\[z=\frac{12}{x}-\frac{1}{4}x\]

Answer 4

volume is
\[V=x^2z=x^2(\frac{12}{x}-\frac{1}{4}x)=12x-\frac{1}{4}x^3\]

Answer 5

take the derivative, set it equal zero and solve. i think you get x = 4 for a max volume of 32 but you should check my arithmetic and algebra

Answer 6

checking with mathematica... impressive

Answer 7

i get x=4, volume=288

Answer 8

wait, no, i think i missed a factor of 3

Answer 9

so what would the dimension of the height be?

Answer 10

it would be x=4, z=2, right?

Answer 11

ok we got x = 4 for sure.
and i think we had volume ( if i didn't screw up) as
\[V(x)=12x-\frac{1}{4}x^3\] so we can find \[V(4)\] by substitution

Answer 12

\[V(4)=12\times 4-4^2=48-16=32\] is what i get

Answer 13

yep, but i'm looking for the dimensions. so, z would actually be z=2

Answer 14

yeah I get x=4, z=2, volume=32. Nice job my friend.

Answer 15

so yeah, height would be 2

Answer 16

ty

Answer 17

This is what mathematica says:

In[930]:= Maximize[{x^2 z, 3*(x^2 + 4 x z) == 144, x > 0}, {x, z}]
Out[930]= {32, {x -> 4, z -> 2}}