<s>[INST] If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$ If $ a $ and $ b $ are real numbers that satisfy the equation $ 17 (a ^ 2 + b ^ 2) -30ab-16 = 0 $, the maximum value of the expression$$\sqrt{16a^2+4b^2-16ab-12a+6b+9}$$goes:
a) $1$
b) $3$
c) $5$
d) $7$
e) $9$
Can anyone give a hint?
Attempt:  I tried Variation, Jacobian, Hessian matrices.
 [/INST] Hint
$$16a^2+4b^2-16ab-12a+6b+9=(4a-2b)^2-2(4a-2b)\cdot(3/2)+9=(4a-2b-3/2)^2+9+(3/2)^2$$
Let $4a-2b=2c\implies b=2a-c$
From the given condition,
$$0=17a^2+17(2a-c)^2-30a(2a-c)-16=a^2(17+68-60)+a(60c-68c)+17c^2-16=0$$
which is a quadratic equation in $a$
As $a$ is real, the discriminant must be $\ge0$
Use this fact to find the range of values of $c$
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