$ y = \frac{8}{\cos^2 x} + 18 \cdot\cos^2 x$ find the minimum possible value of y.

Question

Given , $y = 8 \sec^2 x + 18 \cos^2 x$, find the minimum possible value of $y$.

Well, basically, it's a Question from Minimizing and Maximizing i.e. (Optimization) Section.

So, my try went as follows -

$ y = \frac{8}{\cos^2 x} + 18 \cdot\cos^2 x$

So, we we can see the value of cos function can't be $90$ degrees or $\frac{\pi}{2}$

Now, comes this part where I'm completely clueless,

If $x$ doesn't equal to $\frac{\pi}{2}$, then it must be reduced such that all functions change to cosine or sec, which here it seems impossible.

Is there a way to do it?

Answer 1

Recall that by AM-GM inequality

$$\frac{8}{\cos^2 x} + 18 \cos^2 x\ge 2\sqrt{144}=24$$

with equality for

$$\frac{8}{\cos^2 x} = 18 \cos^2 x$$

Answer 2

$$y = \frac{8}{\cos^2 x} + 18\cos^2 x= \left( \sqrt{18} \cos x -\frac{\sqrt{8}}{\cos x}\right)^2+24 \ge 24$$

Answer 3

Using calculus: $$\begin{align} y &= \frac{8}{\cos^2 x} + 18 \cdot\cos^2 x=8+8\tan^2x+18\cos^2x;\\ y'&=\frac{16\tan x}{\cos^2x}-36\cos x\sin x=0 \Rightarrow 16\sin x=36\cos^4x\sin x \Rightarrow \cos^2x=\frac23;\\ y''&=\frac{16+32\sin^2 x}{\cos^4x}+36\sin^2x-36\cos^2x=\\ & \ \ \ \ \ \frac{48-32\cos^2 x}{\cos^4x}+36-72\cos^2x=\\ & \ \ \ \ \ 60+36-48=48>0\end{align}$$ Hence: $$y_{min}=\frac{8}{2/3}+18\cdot \frac23=24.$$