Maximizing $a+b$ given Quadratic Constraint

[anemone]

Find the maximum of $a+b$, given $a^2-1+b^2-3b=0$.

 

[MarkFL]

My solution:

Spoiler

Using Lagrange multipliers, we obtain:

$$1=\lambda(2a)$$

$$1=\lambda(2b-3)$$

This implies:

$$a=b-\frac{3}{2}$$

Substituting into the constraint, we obtain:

$$\left(b-\frac{3}{2} \right)^2-1+b^2-3b=0$$

$$2b^2-6b+\frac{5}{4}=0$$

$$8b^2-24b+5=0$$

$$b=\frac{6\pm\sqrt{26}}{4}$$

Hence:

$$a=\frac{\pm\sqrt{26}}{4}$$

And so the maximum of the objective function $f(a,b)=a+b$ is:

$$f_{\max}=f\left(\frac{\sqrt{26}}{4},\frac{6+\sqrt{26}}{4} \right)=\frac{3+\sqrt{26}}{2}$$

 

[anemone]

My solution:

Spoiler

Using Lagrange multipliers, we obtain:

$$1=\lambda(2a)$$

$$1=\lambda(2b-3)$$

This implies:

$$a=b-\frac{3}{2}$$

Substituting into the constraint, we obtain:

$$\left(b-\frac{3}{2} \right)^2-1+b^2-3b=0$$

$$2b^2-6b+\frac{5}{4}=0$$

$$8b^2-24b+5=0$$

$$b=\frac{6\pm\sqrt{26}}{4}$$

Hence:

$$a=\frac{\pm\sqrt{26}}{4}$$

And so the maximum of the objective function $f(a,b)=a+b$ is:

$$f_{\max}=f\left(\frac{\sqrt{26}}{4},\frac{6+\sqrt{26}}{4} \right)=\frac{3+\sqrt{26}}{2}$$

Bravo, my dearest admin!(Party) And thanks for participating!

 

[anemone]

Solution suggested by other:

Spoiler

$a^2-1+b^2-3b=0$ can be rewritten as $a^2+\left( b-\dfrac{3}{2} \right)^2=\dfrac{13}{4}$ and this represents an equation of a circle center at $\left( 0,\,\dfrac{3}{2} \right)$ with radius $\dfrac{\sqrt{13}}{2}$ and also its parametric equation as $a=\dfrac{\sqrt{13}}{2}\cos \theta$ and $b=\dfrac{\sqrt{13}}{2}\sin \theta+\dfrac{3}{2}$.

Thus $a+b=\dfrac{3}{2}+\dfrac{\sqrt{13}}{2}(\cos \theta+\sin \theta)=\dfrac{3}{2}+\dfrac{\sqrt{13}}{2}(\sqrt{2}\sin (\theta+45^{\circ})$ which is maximized when $\theta=45^{\circ}$ and gives the maximum of $a+b$ as $\dfrac{\sqrt{26}+3}{2}$.