Possible typo in problem of Braun's Differential Equations

Question

In Martin Braun's book Differential Equations, problem 16 in section 1.10 says:

Consider the initial-value problem $y'= t^2+y^2$, $y(0)=0$ ... (*)

and Iet $R$ be the rectangle $0\leq t \leq a, - b \leq y \leq b$

(a) Show that the solution $y(t)$ of (*) exists for $0 \leq t \leq min(a, b/(a^2+b^2))$

(b) Show that the maximum value of $b/(a^2 +b^2)$, for a fixed $a$, is $1/2a$.

(c) Show that $\alpha=min(a,\frac{1}{2}a)$ is largest when $a=\frac{1}{2^{1/2}}$.

(d) Conclude that the solution $y ( t)$ of (*) exists for $0 \leq t \leq \frac{1}{2^{1/2}} $

The problem is, in the incise b) i proof by calculus of the maximum of the function $f(b)=\frac{a^2}{a^2+b^2}$, the maximun is reached when $f(a)=\dfrac{1}{2a}$, but in incise c) the book puts $\alpha=min(a,\frac{1}{2}a)$, my question is, that is an error? or i'm wrong? it should be $\alpha=min(a,\frac{1}{2a})$? or not? and, how can i prove that? thanks and, i apologize if the question is clumsy, but i tried to solve this problem for hours and the book confused me, because the notation.

Answer 1

Indeed, that is clearly a misprint in c). The $\frac{1}{2}a$ should be $\frac{1}{2a}$, so that c) reads

c) Show that $\alpha=\min(a,1/(2a))$ is largest when $a=1/\sqrt{2}$.

Good luck with your further studies!