We are asked to find the minimum cost of a box subject to the following constraints: the volume is 70 cubic meters, the length is twice the width for the top and bottom, the cost of the material for the top and bottom is $9 per square meter, and the cost for the side material is $6 per square meter.
Let w be the width of the bottom, so 2w is the length, and h is the height of the box. The surface area of the box is `SA=2(w)(2w)+2(h)(2w)+2(h)(w)`
or `SA=4w^2+6hw`
Then, the cost of the materials will be `C=4w^2(9)+6(6hw)=36w^2+36hw`
We would like to have this in terms of a single variable. We use the fact that the volume is 70 to write C in terms of w:
`70=w(2w)(h) => h=35/w^2`
So, `C=36w^2+1260/w`
We can use calculus to find the minimum. The domain of the function C(w) is positive reals, so C is defined on its domain. The minimum, assuming there is one, will occur when the first derivative is zero.
`C'(w)=72w-1260/w^2` If C'=0, then
`72w=1260/w^2 => w^3=17.5 => w~~2.5962`
Then, `l=2w~~5.1925, h=35/w^2~~5.1925`
So, the minimum cost is `C=36w^2+1260/w~~727.9739`
or approximately $727.97
(See attachment for the graph of C.)
A quick check shows
C(1)=1296
C(2)=774
C(3)=744
C(4)=891
C(5)=1152
so, the answer is reasonable. Also `2(2.5962)^2(5.1925)~~69.9975`, so it fits the requirements.
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.
Further Reading