Question

A rectangular storage container with an open top is to have a volume of 10m^3. The lenght of the base is twice the width. material for the base costs 10$ per square meter. material for the sides costs 6$ per square meter. What is the cost of the materials for the cheapest such container

Answer 1

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Answer 2

Hey could you give me just a second while I gather my thoughts

Answer 3

Ok
So start with what is known. The total price will be:
\[P=(10lw)+2(6lh)+2(6wh)\]
Also, we know that the volume is
\[10=lwh\]
and that the length is twice the width
\[l=2w\]

Answer 4

ok so substitute l=2w into 10=lwh
and you get
\[10=2w^2h\]
and simplify for h, this will be important later
\[h=5/w^2\]

Answer 5

Now if we take the price equation and substitute l=2w, we get
\[P=20w^2+24wh+12wh\]
now substitute h=5/w2
and you get
\[P=20w^2+180/w\]

Answer 6

Now comes the tricky part; you can either graph the equation and look for a minimum, or you can take the derivative to find the minimum.
I'll take the derivative:
\[d/dw(20w^2+180/w)=40w-180/w^2\]

Answer 7

So P' is equal to \[40w-180/w^2\]
Now set the equation equal to zero, so that P' is equal to zero (by definition, this shows the minimum and maximum of the equation)

Answer 8

I got a zero at \[w=1.651\]

Answer 9

Now solve for h and you get 5/1.651^2
so h=1.834
and l=3.30193

Answer 10

putting it into the price equation and you get
54.51+72.68+36.34=$163.54

Answer 11

\[P=$163.54\]