Calc Optimization problem with open top A rectangular box with an open top has a volume of 4500 ft^3. The base is made of slate, and the sides are made of glass. Slate is 3 times the price of glass per sq. ft. What dimensions minimize the cost?
 A: Perhaps easier without calculus - we have $l \times b \times h = 4500$.  Also the cost is proportional to $C = 3lb + 2(l+b)h = 3lb + 2lh + 2bh$.  
This is a sum of three terms, which have a constant product.  Hence it gets minimised when the terms are the same, viz. $$3lb = 2lh = 2bh \implies l = b = \frac23 h = \sqrt[3]{\frac{3\times 4500}{2}} = 15\sqrt[3]2$$ 
A: HINT:
If the price of glass per sq. ft.  is $p,\iff$ the price of slate  per sq. ft. $=3p$
If $l,b,h$ be length, breadth and the height (in ft) of the box, $l\cdot b\cdot h=4500$
So, the total price $f(l,b,h)=l\cdot b\cdot3p+2h(l+b)\cdot p$
Now, use Lagrange Multiplier with $g(l,b,h)=l\cdot b\cdot h-4500$
A: We calculate the cost of  a box that has base $x\times x$, and height $h$. Imagine that the cost of glass is $c$ dollars per square foot. Then the cost of slate is $3c$ dollars per square foot.
The area of the glass part is $4xh$. The area of the base is $x^2$. Thus the cost of material for the box is 
$$(c)(4xh)+(3c)(x^2).\tag{1}$$
Note that the volume is $4500$ cubic feet (big box). Thus $x^2h=4500$, and therefore $h=\frac{4500}{x^2}$.
Substituting in (1), and simplifying a bit, we find that the cost $C(x)$ of the box, as a function of $x$, is given by
$$C(x)=c\left(\frac{(4)(4500)}{x}+3x^2\right).$$
Now use the ordinary "calculus" minimization procedure. 
