<s>[INST] maximum value of $ac+bd$ 
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is 

what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please
 [/INST] Let $a = r_1 \cos p, b = r_1 \sin p, c = r_2 \cos q$, and $ d = r_2 \sin q$, where $0 < r_1 ≤ \sqrt{2}$ and $0 < r_2 ≤ 2$. This satisfies the inequalities mentioned in the problem.
Then:
$$ac+bd = r_1 r_2\cos p \cos q + r_1 r_2 \sin p \sin q = r_1 r_2 ( \cos p \cos q + \sin p \sin q) = r_1 r_2 \cos(p-q)$$
Since $\cos x ≤ 1$ for all real $x$, the maximum value is $\sqrt{2} \cdot 2 \cdot 1 = 2 \sqrt{2}$.

Sidenote: This is only one choice of variables. Another choice of variables such as $a = r_1 \sin p, b = r_1 \cos p, c = r_2 \cos q, d = r_2 \sin q$ leads to the same result. Try it yourself!
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