Find $m$ such that area of the set is minimum

Question

Let $m>0$ and $$A=\left\{\left.(x,y)\right|m\leq x\leq 2m, 0\leq y \leq x+\frac 6{x^2}\right\}$$ Find $m$ such that the are of the set is minimum.

I used the double integral method to do this but however can I do it without it? since this is a 12th grade question and we don't learn double integrals..

This is what I did, area of the set means that we have:

$$\int_m^{2m}\int_0^{x+\frac 6{x^2}}dydx=\int_m^{2m}(x+\frac 6{x^2})dx=\frac {3(m^3+2)}{2m}.$$

So to minimize the are I set $f(m)= \frac {3(m^3+2)}{2m}$ and used derivatives to find the point of minima..

$$f'(m)=\frac 32(\frac {2m^3-2}{m^2})=0 \to m^3=1$$

Leading to a unique solution $m = 1$.

Can I do this without using the double integral and then just basically solving the question?

Answer 1

By using the AM-GM inequality $$ a+b+c\ge3\sqrt[3]{abc}$$ one has $$ m+\frac{2}{m^2}=\frac12m+\frac12m+\frac{2}{m^2}\ge3\sqrt[3]{\frac12m\cdot\frac12m\cdot\frac{2}{m^2}}=\frac{3\sqrt[3]2}{2} $$ and "=" holds if and only if $\frac12m=\frac12m=\frac{2}{m^2}$ or $m=1$. Thus when $m=1$, the area has the minimum.

Answer 2

If you have some knowledge of basic calculus, you could do it by employing the geometric meaning of integral and chain rule.

Let $$g(x) = x + \frac{6}{x^2},$$ $$A_1 = \left\{(x, y) | 0 \leq x \leq m, 0 \leq y \leq g(x)\right\},$$ $$A_2 = \left\{(x, y) | 0 \leq x \leq 2m, 0 \leq y \leq g(x)\right\},$$ then $S_A(m) = S_{A_2}(m) - S_{A_1}(m)$. With the geometric interpretation of integral, we know that $$\frac{d\, S_{A_1}(m)}{d\,m} = g(m),$$ $$\frac{d\, S_{A_2}(m)}{d\,m} = \frac{d\, S_{A_2}(m)}{d\,(2m)} \cdot\frac{d\, (2m)}{d\,m} = 2g(2m).$$ As a result, the optimal $m$ satisfies $$ \frac{d\, S_{A}(m)}{d\,m} = 2g(2m) - g(m) = 0,$$ which would also lead to the result. Actually, these two methods are equivalent.