Maximum of a function $ f(t)= at+b \sqrt{1-t^2}$

Question

Let $ a,b \in \mathbb{R}$ and let $ f(t)= at+b \sqrt{1-t^2}$ for $t \in [-1,1]$. How do we find the maximum value of $f(t)$ on this range?

Answer 1

By C-S $$at+b\sqrt{1-t^2}\leq\sqrt{(a^2+b^2)(t^2+1-t^2)}=\sqrt{a^2+b^2}.$$ The equality occurs for $(a,b)||(t,\sqrt{1-t^2})$

Answer 2

An idea: $t=sinx$, $f\circ sin=g$

Now we have: $$g(x)=asinx+bcosx,\ x\in(-\pi,\pi)$$

$$g'(x)=acosx-bsinx\\ g'(x)=0 <=>acosx=bsinx <=> tgx=\frac{a}{b} <=> x=arctg\frac{a}{b}$$

$t=sin(arctg\frac{a}{b})$

The problem is that the way this function will behave highly depends on a and b. You can't solve well this problem unless you know more about a and b.