I am currently doing an investigation in which I am required to design the dimensions of a juice box (can be cube/cuboid) which has the least possible surface area that can hold 200 ml of juice.
I have combined these surface area and volume formulas:
$$V = lwh$$
$$A = 2(lw + lh + wh)$$
(V = volume. A = surface area. l = length. w = width. h = height.)
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To make this:
$$A = 2(lw + \frac{200}{w} + \frac{200}{l})$$
(the V was substituted with the volume, 200 ml)
...
So, now that I have an equation (which I don't know if it is correct), I would like to find the least possible value for A, the surface area. If I am not mistaken, I will need to find the derivative of the equation using differential calculus. But, what then? Is that already the least possible surface area?
Apologies, I am new to the concept of calculus, so please inform me if I have done anything wrong in my calculations.
Let us admit that we do not know the answer (that is to say that the smallest surface area of a cuboid for a given volume is a cube as calculus commented).
You properly arrived to $$A = 2\big(lw + \frac{200}{w} + \frac{200}{l}\big)$$ Since you want $A$ to be minimum, you need to set its derivatives with respect to the variables $l,w$ equal to $0$ (think about the bottom of a valley).
So, $$A'_l=2 \left(w-\frac{200}{l^2}\right)=0\tag 1$$ $$A'_w=2 \left(l-\frac{200}{w^2}\right)=0\tag 2$$ From $(1)$, $w=\frac{200}{l^2}$. Plug this in $(2)$ to get $$l-\frac{l^4}{200}=l\big(1-\frac{l^3}{200}\big)=0$$ I suppose that you will accept to forget the case $l=0$. So, what remains is $l^3=200$ which corresponds to $l= \sqrt[3]{200}$. Back to $w$, you will get the same value as well as for $h$.
So, we face a cube of side equal to $ \sqrt[3]{200}\approx 5.84804$.
You could change the specific $200$ to any $V$ and get a cube of side dimension $ \sqrt[3]V$.