I need help solving this math problem:
A rectangular storage container with an open top is to have a volume of 10m3. The length of its base is twice the width. Material for the base costs $10, material for the sides cost 6. Find a function of the cost of material depending on the length of the base side.
I tried solving it
and additional notes
. You'll see, why I didn't try to type it all in.
I found some similar problems on this site and in other places, however the solutions offered don't seem to be applicable to this specific question.
The solution (as presented by the teacher) should be: $C(b)=20b^2+\dfrac{180}{b}$ I (seem to) manage to calculate the first part, however I get $\dfrac{b}{10}$ for the second part (instead of $\dfrac{180}{b}$).
Can anyone please help point me to my mistake.
Let $w$ and $2w$ be the length and breadth respectively and $h$ be the height.
Then the area of base will be $(2w).(w)= 2w^2$. So the cost (in $\$ $) of the material for the base is $10.(2w^2)$. Two of the sides have the area $w.h$ and the other two have $2w.h$; so the cost of the material for the sides is $6[2(2w.h)+2(w.h)]$.
So the total cost is therefore; $$C=10.(2w^2) + 6[2(2w.h)+2(w.h)]=20w^2 + 36wh $$
To express $C$ as a function of $w$ alone we need to eliminate $h$ and that can be done by using given volume.
$$w.2w.h=10$$
Which gives $$h=\frac{5}{w^2}$$
Substitute this in the expression of $C$ and we can easily get $$C(w)= 20w^2 + \frac{180}{w}$$ where $w$ is positive.