How to find the greatest value of this expression?

Question

"This expression" is

$$\frac{5\sin^2\alpha+4\cos^2\alpha}{4\cos^2\beta+5\sin^2\beta}.$$ The answer is $1.25$

I used simple steps to simplify this, but couldn't find the greatest value, since it has $2$ kinds of angles. So how to find that?

Answer 1

Note also that you can write your expression as $$\frac{\sin^2\alpha + 4(\overbrace{\sin^2\alpha + \cos^2\alpha}^1)}{\sin^2\beta + 4(\underbrace{\sin^2\beta + \cos^2\beta}_1)}$$ $$=\frac{4+ \sin^2\alpha}{4+ \sin^2\beta}$$

A ratio of positive numbers increases if the numerator increases, or if the denominator decreases. Since $0\leq\sin^2\theta\leq 1$, this means that the largest value of the numerator is $4+1=5$ and the smallest value of the denominator is $4+0=4$. So the largest value of the fraction is $\frac54$.

Answer 2

Hint. Maximize the numerator and minimize the denominator (they depends on different variables). As regards the numerator note that $$n(\alpha):=5\sin^2\alpha+4\cos^2\alpha=\sin^2+4\alpha\leq 5=n(\pi/2).$$ Can you handle the denominator?

Answer 3

since numerator and denominator depend on different parameters maximize the numerator and minimize the denominator. $$ \alpha=\pi/2; \,\,\,\,\,\,\beta=0 $$ to get $5/4$ as maximum value

Answer 4

$$\frac { 5\sin ^{ 2 } \alpha +4\cos ^{ 2 } \alpha }{ 4\cos ^{ 2 } \beta +5\sin ^{ 2 } \beta } =\frac { \sin ^{ 2 } \alpha +4 }{ 4+\sin ^{ 2 } \beta } =\frac { 1+4 }{ 4 } =1.25$$

Answer 5

Hint: $\mathrm{max}\{ 5\sin^2\alpha + 4\cos^2\alpha \} = 5$ and $\mathrm{min}\{ 4\cos^2\beta + 5\sin^2\beta \} = 4$

Answer 6

HINT: your term is equivalent to $${\frac { \left( \cos \left( \alpha \right) \right) ^{2}-5}{ \left( \cos \left( \beta \right) \right) ^{2}-5}} $$

Answer 7

Setting $$ x=\cos^2\alpha,\quad y=\cos^2\beta, $$ we have $$ \dfrac{5\sin^2\alpha+4\cos^2\alpha}{4\cos^2\beta+5\sin^2\beta}=\dfrac{5(1-x)+4x}{4y+5(1-y)}=\dfrac{5-x}{5-y}=f(x,y) $$ since $$ 0\le x,y\le 1, $$ we have $$ 4\le 5-x,5-y\le 5 $$ and therefore $$ \dfrac{4}{5}\le f(x,y)\le \dfrac{5}{4}. $$ Notice that $$ f(0,1)=\dfrac{5}{4}, $$ thus $$ \max_{\alpha,\beta}\dfrac{5\sin^2\alpha+4\cos^2\alpha}{4\cos^2\beta+5\sin^2\beta}=\max_{0\le x,y\le 1}f(x,y)=f(0,1)=\dfrac54=1.25 $$

Answer 8

$$\frac{5\sin^2\alpha+4\cos^2\alpha}{4\cos^2\beta+5\sin^2\beta}=\frac{4+\sin^2\alpha}{4+\sin^2\beta}\leq\frac{4+1}{4}=\frac{5}{4}.$$ The equality occurs for $\sin\alpha=1$ and $\sin\beta=0$, which says that the answer is $\frac{5}{4}.$

Done!