Duncan Reed

Answered

2022-01-26

If $u=\sqrt{a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x}+\sqrt{b{\mathrm{cos}}^{2}x+a{\mathrm{sin}}^{2}x}$, find the maximum and minimum value of ${u}^{2}$.

Answer & Explanation

Devyn Figueroa

Expert

2022-01-27Added 10 answers

$u=\sqrt{a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x}+\sqrt{b{\mathrm{cos}}^{2}x+a{\mathrm{sin}}^{2}x}$
Let
$p=a{\mathrm{cos}}^{2}x+b{\mathrm{sin}}^{2}x$
$q=b{\mathrm{cos}}^{2}x+a{\mathrm{sin}}^{2}x$
and
$u=\sqrt{p}+\sqrt{q}$
Then ${u}^{2}=p+q+2\sqrt{pq}$
Now
p+q=a+b
and
$pq=\frac{{\left(a+b\right)}^{2}}{4}-\frac{{\left(a-b\right)}^{2}}{8}-\frac{{\left(a-b\right)}^{2}}{8}\mathrm{cos}4x$
If $\mathrm{cos}4x=-1$ then ${u}^{2}$ is maximum and is equal to
2(a+b)
If $\mathrm{cos}4x=1$ then ${u}^{2}$ is minimum and is equal to
$a+b+2\sqrt{ab}$

Armani Dyer

Expert

2022-01-28Added 10 answers

Put
$1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}x$
$1+\mathrm{cos}2x=2{\mathrm{cos}}^{2}x$

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