If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$

Question

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.

Answer 1

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 $c$

Answer 2

The hint.

Let $2a-b=k.$

Thus, we need to find a maximal value of $\sqrt{4k^2-6k+9}.$

Now, find all values of $k$, for which the equation $$k^2(17(a^2+b^2)-30ab)=16(2a-b)^2$$ has solutions and choose from these values such that $\sqrt{4k^2-6k+9}$ will get a maximal value.

I got that it happens for $k=-\frac{5}{2}.$