A rectangular storage container with an open top is to have a volume of $70$ cubic meters. The length of its base is twice the width. Material for the base costs $70$ dollars per square meter. Material for the sides costs $5$ dollars per square meter. Find the cost of materials for the cheapest such container.
I know that $$v=LWH$$
$$L=2W$$
$$C(cost)=70LW+5(2LH+2WH)=70LW+10LH+10WH$$
$$V=(2W)WH=2HW^2$$
$$2HW^2=70$$
$$C=70(2W)W+10(2W)H+10(WH)=140W^2+30WH$$
$$H=70/2W^2=35/W^2$$
$$C=140W^2+30WH=140W^2+30W(35/w^2)=140W^2+1050W/W^2$$
$$C=140W^2+1050/W$$
So now I would need to find the minimum of the function, but this is where I'm having trouble.
[Math] Minimum cost of a rectangle storage container
calculus
Best Answer
$C(w) = 140w^2 + \dfrac{1050}{w}$, then $C'(w) = 280w - \dfrac{1050}{w^2} = 0 \to w = \sqrt[3]{\frac{1050}{280}}$. This $w$ gives the lowest cost. You can find the rest using this $w$.