If a > 0, b > 0 and 2a + b = 1, find the minimum value of a/(2 − 2a) + b/(2 − b).

Step-by-step solution

1. 1

   Reduce to one variable and pin down the domain. The constraint gives $b=1-2a$, and $a>0,\ b>0$ forces

   $0<a<\tfrac12$

   On that interval $2-2a\in(1,2)$ and $2-b=1+2a\in(1,2)$, so both denominators stay positive and $T$ is continuous. Substituting,

   $T(a)=\frac{a}{2(1-a)}+\frac{1-2a}{1+2a}$

2. 2

   Differentiate each piece separately. For the first term, the quotient rule gives a numerator $1\cdot 2(1-a)-a\cdot(-2)=2$, so

   $\frac{d}{da}\left[\frac{a}{2(1-a)}\right]=\frac{2}{4(1-a)^2}=\frac{1}{2(1-a)^2}$

   For the second, the numerator is $(-2)(1+2a)-(1-2a)(2)=-4$, so

   $\frac{d}{da}\left[\frac{1-2a}{1+2a}\right]=\frac{-4}{(1+2a)^2}$

   The first term is increasing and the second is decreasing, which is exactly why an interior minimum exists.

3. 3

   Set the derivative to zero and solve.

   $\frac{1}{2(1-a)^2}=\frac{4}{(1+2a)^2}\quad\Longrightarrow\quad (1+2a)^2=8(1-a)^2$

   On $0<a<\frac12$ both $1+2a$ and $1-a$ are positive, so take positive square roots:

   $1+2a=2\sqrt{2}\,(1-a)\quad\Longrightarrow\quad a\left(2+2\sqrt{2}\right)=2\sqrt{2}-1$

   $a=\frac{2\sqrt{2}-1}{2+2\sqrt{2}}=\frac{5-3\sqrt{2}}{2}\approx 0.378680$

   (the last step multiplies top and bottom by $\sqrt{2}-1$, using $(1+\sqrt{2})(\sqrt{2}-1)=1$). Then $b=1-2a=3\sqrt{2}-4\approx 0.242641>0$, so the critical point is admissible.

4. 4

   Evaluate each term in exact form. With $1-a=\frac{3(\sqrt{2}-1)}{2}$,

   $\frac{a}{2-2a}=\frac{5-3\sqrt{2}}{6(\sqrt{2}-1)}=\frac{(5-3\sqrt{2})(\sqrt{2}+1)}{6}=\frac{2\sqrt{2}-1}{6}$

   and with $2-b=6-3\sqrt{2}$,

   $\frac{b}{2-b}=\frac{3\sqrt{2}-4}{3(2-\sqrt{2})}=\frac{(3\sqrt{2}-4)(2+\sqrt{2})}{6}=\frac{\sqrt{2}-1}{3}$

5. 5

   Add them and confirm it is the minimum.

   $T_{\min}=\frac{2\sqrt{2}-1}{6}+\frac{2\sqrt{2}-2}{6}=\frac{4\sqrt{2}-3}{6}\approx 0.442809$

   Compare the endpoints of the open interval: as $a\to 0^{+}$, $T\to 1$; as $a\to\tfrac12^{-}$, $T\to\tfrac12$. Both exceed $0.4428$, so the interior critical point is the global minimum.

6. 6

   Watch out for two tempting wrong answers. Plugging in the symmetric-looking $a=b=\frac13$ gives $\frac14+\frac15=\frac{9}{20}=0.45$, which is close but not minimal — $a=b$ does not satisfy the weighted constraint in a balanced way. And the value $\frac15=0.2$ is impossible here, since $T$ never drops below $\frac{4\sqrt{2}-3}{6}\approx 0.4428$ anywhere on $0<a<\frac12$. The exact minimum is $\frac{4\sqrt{2}-3}{6}$.