Max Value of $a^2+2a+b^2$ When $2a^2-6a+b^2=0$

[anemone]

If the real numbers $a$ and $b$ satisfy the condition $2a^2-6a+b^2=0$, find the maximal value of $a^2+2a+b^2$.

 

[kaliprasad]

If the real numbers $a$ and $b$ satisfy the condition $2a^2-6a+b^2=0$, find the maximal value of $a^2+2a+b^2$.

Spoiler

we need to maximize $a^2+2a+b^2$ or $a^2+2a+b^2-(2a^2-6a+b^2)$ as $2^{nd}$ expression is zero hence constant
or $8a-a^2 = 16-(a^2-8a + 16) = 16-(a-4)^2$
above is maximum when $|a-4|$ minimum.
we need to find it under given constraint
which is
$2a^2 -6a + b^2 = 0$
or $(4a^2 - 12a + 9) + 2b^2 = 9$
or $(2a-3)^2 + b^2 = 9$
or $- 3 < = 2a - 3 <= 3$ or a has to be between 0 and 3
so a = 3 and maximum value of given expression = 15.

 

[MarkFL]

If the real numbers $a$ and $b$ satisfy the condition $2a^2-6a+b^2=0$, find the maximal value of $a^2+2a+b^2$.

My solution:

Spoiler

In order for $a$ to be real, we require:

$$(-6)^2-4(2)(b^2)\ge0$$

$$9-2b^2\ge0$$

$$0\le b^2\le\frac{9}{2}$$

In order for $b$ to be real, we require:

$$6a-2a^2\ge0$$

$$a(3-a)\ge0\implies 0\le a\le3$$

For simplicity, let's write the objective function in terms of $a$ only using the constraint:

$$f(a)=a^2+2a+6a-2a^2=8a-a^2$$

$$f'(a)=8-2a=2(4-a)$$

On the domain for $a$, we find $f$ is strictly increasing, hence:

$$f_{\max}=f(3)=15$$

 

[anemone]

Very good job to both of you! And thanks for participating!(Cool)(Wink)