Question

Calculus 1: Optimization
A rectangular storage container with an open top has a volume of 10cm^3. The length of the base is to be twice the width of the base. Find the dimensions of the length, width, and height that minimize the box's surface area. Keep in mind that the top is open.

Answer 1

volume of box = 2x*x*h = 10
2x^2h = 10
x^2 h = 5
surface area = 2x^2 + 2xh + 4xh
= 2x^2 + 6xh

now plug in y = 5 / x^2 form volume formula into area formula
find dA/dx and equate to zero to find x

Answer 2

it says " The length of the base is to be twice the width of the base."
I assume that is supposed to be base is twice height, or..?

Answer 3

oh I see now

Answer 4

What I figured on that was that it is the base of the box, i.e., a rectangle. So the area of that rectangle would be w*(2w)

Answer 5

nevermind :)

Answer 6

then crwr is right

Answer 7

I got up to the plugging into the SA part, but now I'm lost.
Shouldn't it be: 2(wh + 2w*w + 2w*h)
2(wh + 2w^2 + 2wh)
2wh + 4w^2 + 4wh
4w^2 + 6wh

and if that is correct, what exactly do I do next?

Answer 8

A = 2x^2 + 6xh
= 2x^2 + 6x * 5
-
x^2

A = 2x^2 + 30 / x

dA/dx = 4x - 30/x^2 = 0
4x^3 - 30 = 0
x^3 = 7.5
x = cube root 7.5 only positive root makes sense

Answer 9

your surface area calculation is contradictory to mine. how did you get just 2x^2, if the width is x, and the length is 2x as defined?

Answer 10

the top is open so its only one 2x *x

Answer 11

I see now

Answer 12

right - and to confirm its a minimum find second derivative
thats 4 - 60/x^3 and plug in value of x - if its positive that confirms ita a minimum