Oh yeah i found the B and H = 7,93700526
L = 2 x 7,93700526
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.
L=2x, B=x, H=h, so volume V= 2x^2 h =1000
K= 2(L.H) + 2(B.H) + 2(L.B) = 4x.h + 2x.h + 2x^2 = 6x.h + 2x^2We 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)
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 . . .