Thread: Calculus Maximize Help

Hi guys i hav an urgent prob.

I have to hand in a container of 1 litre on monday, but im not sure if my calculations r correct.

The container shud have a length twice the width.
It shud have a vol of 1000 ml
It shud hav a minimum exterior area.

Ok this is what i did.

L = 2x
B = x
H = x
Vol = L . B . H

1000 ml = 2x^3

We dont have to find the derivative if we hav a max value for the volume already or do we?


K, the exterior area:
Ex-Area = (2L+B)H + 2(LxB)
In the end the Ex Area = 10x^2

But what now? Shud we find the derivative?
Pls help me.

Oh yeah i found the B and H = 7,93700526
L = 2 x 7,93700526

Originally Posted by janvdl

Hi guys i hav an urgent prob.

I have to hand in a container of 1 litre on monday, but im not sure if my calculations r correct.

The container shud have a length twice the width.
It shud have a vol of 1000 ml
It shud hav a minimum exterior area.

Ok this is what i did.

L = 2x
B = x
H = x
Vol = L . B . H

1000 ml = 2x^3

L=2x, B=x, H=h, so volume V= 2x^2 h =1000

We dont have to find the derivative if we hav a max value for the volume already or do we?


K, the exterior area:
Ex-Area = (2L+B)H + 2(LxB)

K= 2(L.H) + 2(B.H) + 2(L.B) = 4x.h + 2x.h + 2x^2 = 6x.h + 2x^2

but from the known volume we have x.h = 1000/x, so:

K = 6000/x + 2x^2

To maximize K we look for solutions to dK/dx = 0.

dK/dx = -6000/x^2 + 4x

so we want solutions of -6000/x^2 + 4x=0, or 6000=4x^3, so

x=cuberoot(1500) ~= 11.45 cm

RonL

Hello, janvdl!

I suggest you use more familiar variables . . .

The rectangular container should have a length twice the width.
It should have a vol of 1000 ml. .It should have a minimum exterior area.

Let L = length, W = width, H = height of the container.

Since the length is twice the width: .L = 2W

The box looks like this:

Code:

            * - - - - - - - *
          /               / |
        /               /   | H
      * - - - - - - - *     |
      |               |     *
    H |               |   /W
      |               | /
      * - - - - - - - *
             2W

The volume of a "box" is: .V .= .L × W × H

So we have: .V .= .(2W)(W)(H) .= .1000 . → . H .= .500/W² .[1]


Now determine the exterior area of the box.

. . Front and back panels: .2 × (2W)(H) .= .4WH

. . Left and right panels: .2 × (W)(H) .= .2WH

. . Top and bottom panels: .2 × (2W)(W) .= .4W²

Hence, the total exterior area is: .A .= .4W² + 6WH .[2]


Substitute [1] into [2]: .A .= .4W² + 6W(500/W²)

. . and we have: .A .= .4W² + 3000W^{-1}


And that is the function we must minimize . . .