Question

The Volume of a container is given by the equation V=bwh, and the Surface Area is given by the equation SA=bw+2bh+2bw. If you want the height of the container to be equal to the sum of the lenght of the base and width, what is the largest container (volume wise) you can create from 40cm^2 of material? What are the dimensions of the container? (Assume 1 less than or equal to b,w less than or equal to 10 & 2 less than or equal to h less than or equal to 20

Answer 1

hey can you please check the question once again?

Answer 2

thats word for word from the sheet my professor gave me, what doesnt look right? ive been stuck on this problem for two days

Answer 3

SA=bw+2bh+2bw <= Can you check this again?

Answer 4

SA=bw+2bh+2bw..........he said at first there was a typo and it was: SA=bw+2bh+2hw
but then said to leave it the first way

Answer 5

i konw its an optimization problem and i have to get the equation in terms of one variable....and i got from the information that h=b+w

Answer 6

open container/ box

Answer 7

problem is fine as is

Answer 8

since there are two variables (once i plug in h) do i use implicit differentiation?

Answer 9

No, SA=bw+2bh+2bw=3bw+2bh=3bw+2bw+2b^2=5bw+2b^2=40.
=>bw=8-0.4*b^2, w=8/b-0.4*b
Now you want to maximize V=bwh=bw(b+w). Plug in the w above and continue...

Answer 10

for that second to last step: bw=8-0.4b^2.....divide each side by b right? to get
w=(8-0.4b^2)/b could you just got over how its w=8/b-0.4b please?

Answer 11

Correct, w=(8-0.4b^2)/b=8/b-0.4b

Answer 12

ok, so then basically plug that into V=bwh, take the derivative, set it equal to zero and then regular optimization problem from there right?

Answer 13

Yup!

Answer 14

*large sigh of relief* thanks a million!

Answer 15

Glad to help :)