Find the maximum and minimum volumes of a rectangular box whose surface area equals 3000 square cm and whose edge length (sum of lengths of all edges) is 280 cm. Hint: It can be deduced that the box is not a cube, so if x, y, and z are the lengths of the sides, you may want to let x represent a side with x not equal to y not equal to z. Find Maximum and Minimum values and where they occur at?

Question

anonymous · Answer

2(x*y+x*z+y*z)=3000

x*y*z=v

x+y+z=280

2(x*y+x*z+y*z)=3000
x*y+x*z+y*z=1500
x(y+z)+y*z=1500

x*y*z=v
y*z=v/x

x+y+z=280
y+z=280-x

plug those two in we get:
x(280-x)+v/x=1500
cleaning up:

280x-x^2+v/x=1500
multiply all by x so we can get v on the other side:

280x^2-x^3+v=1500x
280x^2-x^3-1500x=v
derive both sides:
560x-3x^2-1500=v'=0

it becomes like 2.7... and 183...

figure out which is max and which is min and youre done. (can do f(2.7), f(183) and those are your max and min)

anonymous · Answer

Jessica Moore that is from yahoo answers and i tried it and it is incorrect.

anonymous · Answer

that 3rd equation from the top is incorrect... it should not be x + y + z = 280... that only adds one of each side... but the question states that 280 is the sum of lengths of ALL edges, not just one each of x, y, and z.

I'm not sure I could work out the rest, but that's one reason the above solution is incorrect...

anonymous · Answer

thank you i got it from there!

anonymous · Answer

great :)  mine was minimal help, but glad it was helpful...