I would say that you are missing a piece of information. Are you sure that you typed the entire problem?
Alright! My final question on here till next semester lol. Doing some last minute review for my final tonight and came across this problem having a little trouble with it.
A rectanglular storage container with an open top is to have a volume of 10 cubic-meters. Material for the sides costs $6 per square-meter. Material for the base costs $10 per square-meter. Find the cost of materials for the cheapest container.
Thanks
{edit} ah yes sorry. Length of its base is twice the width.
L = 2x
W = x
H = h
So Volume = 2x^2h = 10
Exterior Area = 2(L + B)H + (L x B) [Because you do not have a lid]
= 2(2x + x)h + (2x . x)
= 4x^2h + 2x^2
= 6xh + 2x^2
But we know that h = (10)/(2x^2)
= 6x((10)/(2x^2)) + 2x^2
= 60x/2x^2 + 2x^2
= 30/x + 2x^2
dA/dx = -30/x^2 + 4x = 0
-30 = -4x^3
7,5 = x^3
x = 1,957433821
Now you can calculate the rest.
This is your sides:30/x
And this is your bottom lid:2x^2
Let the width be w metres, then the length is 2w metres, and as the volume
is 10 cubic metres the height is: 5/w^2 metres.
Total area of the sides is:
As = 2*w*5/w^2 + 4*w*5/w^2
... = 30/w
The area of the base is:
Ab = 2*w^2
So the cost is:
C = 180/w + 20*w^2
Now you need to find the value of w that minimises C, and that minimum
cost is your answer
RonL
Ok
let x, y, z be the dimensions of the box.
Since volum should be 10, (1) xyz=10
Length of its base is twice the width: (2) y=2z
(3) Cost=6(2xy+2xz)+10yz=12xy+12xz+10yz
You need a function cost having just one variable, so let's use (1) and (2) to simplify (3)
(2) into (1) gives 2xz^2=10 or x=5/z^2
Using this relation and (2) into (3) gives:
cost=20z^2+180/z
Solve the equation Cost'=0 to find z which gives minimum cost.
40z-180/z^2=0
40z^3-180=0
z^3=4.5
z=1.65
x=1.84
y=3.3