Minimize $\frac{2}{1-a}+\frac{75}{10-b}$

Question

Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$

WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.

Consider making a transformation like this

$$\dfrac{2}{1-a}+\dfrac{75}{10-b}=\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}.$$

If $2m=2n=1$, $ma=1-a$, and $ nb=10-b$ can all hold, we can apply AM-GM inequality as follows

$$\frac{2\cdot (ma)^2}{ma\cdot ma\cdot (1-a)}+\frac{75\cdot(nb)^2}{nb\cdot nb\cdot(10-b)}\geq \frac{2\cdot (ma)^2}{\left(\frac{ma +ma+(1-a)}{3}\right)^3}+\frac{75\cdot(nb)^2}{\left(\frac{nb+nb+(10-b)}{3}\right)^3}=\cdots$$ But this is invalid, since $n=\frac{1}{2},nb=10-b$ can not satisfy $b=5$. How to solve it?

Answer 1

I come up with a solution which is based on the clue I posted above.

\begin{align*} \frac{2}{1-a}+\frac{75}{10-b}&=\frac{2}{1-a}+\frac{15}{2}\cdot\frac{b}{10-b}+\frac{15}{2}\\ &=\frac{2\cdot \frac{a}{2}\cdot \frac{a}{2}}{(1-a)\cdot \frac{a}{2}\cdot\frac{a}{2}}+\frac{15}{2}\cdot\frac{b\cdot b}{(10-b)\cdot b}+\frac{15}{2}\\ &\ge\frac{\frac{a^2}{2}}{\left(\frac{(1-a)+\frac{a}{2}+\frac{a}{2}}{3}\right)^3}+\frac{15}{2}\cdot\frac{b^2}{\left(\frac{(10-b)+ b}{2}\right)^2}+\frac{15}{2}\\ &\ge \frac{27}{2}\left(a^2+\frac{b^2}{45}\right)+\frac{15}{2}\\ &=\frac{27}{2}\cdot1+\frac{15}{2}\\ &=21, \end{align*} which hold as equality iff $1-a=\dfrac{a}{2},10-b=b,$ namely $a=\dfrac{2}{3},b=5$.

Answer 2

The Tangent Line method helps.

Let $a=\frac{2}{3}x$ and $b=5y$.

Thus, $$4x^2+5y^2=9$$ and $$\frac{2}{1-a}+\frac{75}{10-b}-21=3\left(\frac{2}{3-2x}+\frac{5}{2-y}-7\right)=$$ $$=3\left(\frac{2}{3-2x}-2-2(x^2-1)+\frac{5}{2-y}-5-\frac{5}{2}(y^2-1)\right)=$$ $$=\frac{6(x-1)^2(2x+1)}{3-2x}+\frac{15y(y-1)^2}{2(2-y)}\geq0.$$ The equality occurs for $x=y=1,$ which says that $21$ is a minimal value.

There is the following reasoning.

Let $$\frac{2}{1-a}+\frac{75}{10-b}=k.$$ Thus, $$b=10-\frac{75}{k-\frac{2}{1-a}}$$ and find all values of $k$ for which the equation $$a^2+\frac{\left(10-\frac{75}{k-\frac{2}{1-a}}\right)^2}{45}=1$$ has solutions $0<a<1.$

Answer 3

There is another way using parameterisation. Let $a = \cos t, b = \sqrt{45} \sin t$, then we have:

$$f(t) = \frac{2}{1-\cos t} + \frac{75}{10-\sqrt{45} \sin t}$$ $$f'(t) = -\frac{2 \sin t}{(1-\cos t)^2} + \frac{225 \sqrt5 \cos t}{(10-3 \sqrt{5} \sin t)^2}$$

The condition that $a,b > 0$ translates to $0 < t < \frac{\pi}{2}$. Setting $f'(t)$ equal to $0$, we find that only $0.841$ (actually $\tan^{-1} \left( \frac{1}{\sqrt5} \right)$) is inside the range.

Therefore the minimum value is $f(0.841) = 21$.

Answer 4

Once we know the minimizer, we can give a simple solution. Here is a way (by hand) to obtain the minimizer (which motivates the simple solution).

Since $\frac{2}{1-a} > 2$ and $\frac{75}{10-b} > \frac{75}{10}$, let $$x := \frac{2}{1-a} - 2 > 0, \quad y := \frac{75}{10-b} - \frac{75}{10} > 0.$$ We have $$a = \frac{x}{2+x}, \quad b = \frac{20y}{15 + 2y}.$$

From $a^2 + \frac{b^2}{45} = 1$, we have $$\frac{x^2}{(2+x)^2} + \frac{1}{45}\frac{(20y)^2}{(15 + 2y)^2} = 1$$ which results in $$y = \frac{135x + 135 + 45(2 + x)\sqrt{5(1+x)}}{10x^2 + 22x + 22}.$$

We have $$\frac{2}{1-a} + \frac{75}{10-b} = 2 + x + \frac{75}{10} + y$$ $$= \frac{19}{2} + x + \frac{135x + 135 + 45(2 + x)\sqrt{5(1+x)}}{10x^2 + 22x + 22}.$$

Let us eliminate the sqrt. Let $u := \sqrt{5(1+x)} > \sqrt{5}$. We have $x = \frac{u^2}{5} - 1$. We have $$\frac{2}{1-a} + \frac{75}{10-b} = \frac{17}2 + \frac{u^2}{5} + \frac{45u}{2u^2 - 6u + 10} =: f(u).$$ We turn to find the minimum of $f(u)$ subject to $u > \sqrt{5}$. We have $$f'(u) = \frac{(u-5)(4u^4-4u^3+56u^2-65u-225)}{10(u^2-3u+5)^2}.$$ (Note: Here we use the rational root theorem to obtain the factor $u - 5$.) We have $f(5) = 21$. $21$ is the minimum since $$f(u) - 21 = \frac{(2u^2 + 14u - 25)(u-5)^2}{10(u^2-3u+5)} \ge 0, \, \forall u > \sqrt{5}.$$ (Note: Here we use the fact that $f(u) - 21$ has a factor $u - 5$.)

Answer 5

The method of Lagrange Multipliers is also one new insight to this question

Let the Lagrangian be: $$L(a, b, \lambda) = \frac{2}{1-a} + \frac{75}{10-b} + \lambda\left(a^2 + \frac{b^2}{45} - 1\right)$$

We take the partial derivatives with respect to $a$ and $b$ and set them to zero:

1. $\frac{\partial L}{\partial a} = \frac{2}{(1-a)^2} + 2a\lambda = 0 \implies \lambda = -\frac{1}{a(1-a)^2}$
2. $\frac{\partial L}{\partial b} = \frac{75}{(10-b)^2} + \frac{2b}{45}\lambda = 0 \implies \lambda = -\frac{75 \cdot 45}{2b(10-b)^2} = -\frac{3375}{2b(10-b)^2}$

Equating the two expressions for $\lambda$: $$\frac{1}{a(1-a)^2} = \frac{3375}{2b(10-b)^2}$$ $$2b(10-b)^2 = 3375 a(1-a)^2$$

Verify if there are simple rational solutions for $a$ and $b$. Let's test $a = \frac{2}{3}$. If $a = \frac{2}{3}$, then substitute into the constraint equation: $$\left(\frac{2}{3}\right)^2 + \frac{b^2}{45} = 1$$ $$\frac{4}{9} + \frac{b^2}{45} = 1$$ $$\frac{b^2}{45} = 1 - \frac{4}{9} = \frac{5}{9}$$ $$b^2 = 45 \cdot \frac{5}{9} = 25 \implies b = 5 \quad (\text{since } b > 0)$$

Now we check if these values satisfy the Lagrange condition $2b(10-b)^2 = 3375 a(1-a)^2$:

$LHS:$ $$2(5)(10-5)^2 = 10(5)^2 = 10(25) = 250$$ $RHS:$ $$3375 \left(\frac{2}{3}\right) \left(1-\frac{2}{3}\right)^2 = 3375 \cdot \frac{2}{3} \cdot \left(\frac{1}{3}\right)^2$$ $$= 3375 \cdot \frac{2}{3} \cdot \frac{1}{9} = \frac{3375 \cdot 2}{27} = 125 \cdot 2 = 250$$

Since LHS = RHS, the point $(a, b) = (\frac{2}{3}, 5)$ is a critical point.

Finally, we evaluate the expression $S$ at this point: $$S = \frac{2}{1-2/3} + \frac{75}{10-5}$$ $$S = \frac{2}{1/3} + \frac{75}{5}$$ $$Ans = S = 6 + 15 = \boxed{21}$$

Analysis of the function behavior at the boundaries ($a \to 0, a \to 1$) and other critical points confirms this is the global minimum.