Determining $(a,b) \in \mathbb{R^2}$ with $a, b \gt 0$ on $f(x)=4-x^2$ so that the triangle between $f(x)$ and the x- and y-axis has minimal area

Question

How can I determine the point $(a,b) \in \mathbb{R^2}$ with $a, b \gt 0$ on $f(x)=4-x^2$ so that the triangle between $f(x)$ and the x- and y-axis has minimal area? I guess it would be possible using derivation but I don't know how I should approach this problem.

Answer 1

Let $f(x) = 4 - x^2$. Notice that if the point $(a,b)$ is in $f$, then $b = f(a)$. In order to have $b > 0$, we must have $f(a) > 0 \Rightarrow 4 - a^2 > 0 \Rightarrow 0 < a < 2$ (since you are also asking for $a > 0$). This in turn gives $0 < b < 4$.

Assuming that you meant that your $3$ points on the triangle are $(a,0), (0,f(a)),$ and $(a,f(a))$, you first need to come up with an expression for the area (call it $A(a)$) in terms of where you choose your $x$-value $a$ to be, and then attempt to minimize it. As a hint, you should get $A(a) = ab/2$. Can you take it from here?

EDIT: Make sure you also read the problem again to verify that it's the minimum area you want.. A triangle like the one described above will have zero area at $a = 0$ and $a = 2$, so it's likely that you actually want the maximum area.