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

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The discussion focuses on maximizing the expression $a^2 + 2a + b^2$ under the constraint $2a^2 - 6a + b^2 = 0$. Participants explore various methods to solve the problem, including substitution and optimization techniques. The condition leads to a quadratic equation in terms of $a$, which can be analyzed to find the maximum value. The conversation emphasizes the importance of correctly applying mathematical principles to derive the solution. Ultimately, the goal is to determine the highest achievable value of the expression given the specified constraint.
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If the real numbers $a$ and $b$ satisfy the condition $2a^2-6a+b^2=0$, find the maximal value of $a^2+2a+b^2$.
 
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anemone said:
If the real numbers $a$ and $b$ satisfy the condition $2a^2-6a+b^2=0$, find the maximal value of $a^2+2a+b^2$.

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.
 
anemone said:
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:

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:

$$f_{\max}=f(3)=15$$
 
Very good job to both of you! And thanks for participating!(Cool)(Wink)
 
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