Question

If a, b > 0 such that a² + 2b² = 12, then-

• A. maximum value of a + 2b is 6
• B. maximum value of ab² is 8
• C. maximum value of $\frac{a}{2} + \frac{b}{4}$ is 3
• D. To get maximum

Answer 1

Answer: (A), (B)

Options (A) and (B) are correct.

Answer 2

Part (A): Maximum value of a + 2b

Let the expression be $E_1 = a + 2b$.
We can use the Cauchy-Schwarz inequality.
Consider the terms $a$ and $2b$. We have the constraint $a^2 + 2b^2 = 12$.
Rewrite $2b$ as $\sqrt{2} \cdot \sqrt{2}b$.
Applying Cauchy-Schwarz inequality:
$(x_1y_1 + x_2y_2)^2 \le (x_1^2 + x_2^2)(y_1^2 + y_2^2)$
Let $x_1 = 1$, $y_1 = a$.
Let $x_2 = \sqrt{2}$, $y_2 = \sqrt{2}b$.
Then $(1 \cdot a + \sqrt{2} \cdot \sqrt{2}b)^2 \le (1^2 + (\sqrt{2})^2)(a^2 + (\sqrt{2}b)^2)$
$(a + 2b)^2 \le (1 + 2)(a^2 + 2b^2)$
$(a + 2b)^2 \le 3 \cdot 12$
$(a + 2b)^2 \le 36$
Since $a, b > 0$, $a+2b > 0$.
So, $a + 2b \le \sqrt{36}$
$a + 2b \le 6$.
The maximum value is 6.
Equality holds when $\frac{y_1}{x_1} = \frac{y_2}{x_2}$, i.e., $\frac{a}{1} = \frac{\sqrt{2}b}{\sqrt{2}}$, which simplifies to $a = b$.
Substitute $a=b$ into the constraint $a^2 + 2b^2 = 12$:
$b^2 + 2b^2 = 12$
$3b^2 = 12$
$b^2 = 4$
Since $b > 0$, $b = 2$.
Then $a = 2$.
Check: $a^2 + 2b^2 = 2^2 + 2(2^2) = 4 + 8 = 12$.
For these values, $a + 2b = 2 + 2(2) = 6$.
Thus, option (A) is correct.

Part (B): Maximum value of ab²

Let the expression be $E_2 = ab^2$.
We can use the AM-GM inequality.
We have $a^2 + 2b^2 = 12$. We want to maximize $ab^2$.
Notice that $ab^2 = \sqrt{a^2 b^4}$.
We can split $2b^2$ into two equal parts: $b^2$ and $b^2$.
Consider the three terms $a^2$, $b^2$, and $b^2$.
Their sum is $a^2 + b^2 + b^2 = a^2 + 2b^2 = 12$.
By AM-GM inequality:
$\frac{a^2 + b^2 + b^2}{3} \ge \sqrt[3]{a^2 \cdot b^2 \cdot b^2}$
$\frac{12}{3} \ge \sqrt[3]{a^2b^4}$
$4 \ge \sqrt[3]{a^2b^4}$
Cube both sides:
$4^3 \ge a^2b^4$
$64 \ge a^2b^4$
Since $a, b > 0$, $ab^2 > 0$.
So, $ab^2 \le \sqrt{64}$
$ab^2 \le 8$.
The maximum value is 8.
Equality holds when $a^2 = b^2 = b^2$, which implies $a^2 = b^2$. Since $a, b > 0$, this means $a = b$.
Substitute $a=b$ into the constraint $a^2 + 2b^2 = 12$:
$b^2 + 2b^2 = 12$
$3b^2 = 12$
$b^2 = 4$
Since $b > 0$, $b = 2$.
Then $a = 2$.
Check: $a^2 + 2b^2 = 2^2 + 2(2^2) = 4 + 8 = 12$.
For these values, $ab^2 = 2 \cdot (2^2) = 2 \cdot 4 = 8$.
Thus, option (B) is correct.

Part (C): Maximum value of $\frac{a}{2} + \frac{b}{4}$

Let the expression be $E_3 = \frac{a}{2} + \frac{b}{4}$.
We can use the Cauchy-Schwarz inequality.
We have $a^2 + 2b^2 = 12$.
Rewrite the expression as $\frac{1}{2}a + \frac{1}{4}b$.
Consider $a = x$ and $\sqrt{2}b = y$. Then $x^2 + y^2 = 12$.
Also, $b = \frac{y}{\sqrt{2}}$.
So, $E_3 = \frac{x}{2} + \frac{y}{4\sqrt{2}}$.
Applying Cauchy-Schwarz inequality:
$(\frac{x}{2} + \frac{y}{4\sqrt{2}})^2 \le ((\frac{1}{2})^2 + (\frac{1}{4\sqrt{2}})^2)(x^2 + y^2)$
$(\frac{a}{2} + \frac{b}{4})^2 \le (\frac{1}{4} + \frac{1}{16 \cdot 2})(a^2 + 2b^2)$
$(\frac{a}{2} + \frac{b}{4})^2 \le (\frac{1}{4} + \frac{1}{32})(12)$
$(\frac{a}{2} + \frac{b}{4})^2 \le (\frac{8+1}{32})(12)$
$(\frac{a}{2} + \frac{b}{4})^2 \le \frac{9}{32} \cdot 12$
$(\frac{a}{2} + \frac{b}{4})^2 \le \frac{9 \cdot 3}{8}$
$(\frac{a}{2} + \frac{b}{4})^2 \le \frac{27}{8}$
Since $a, b > 0$, $\frac{a}{2} + \frac{b}{4} > 0$.
So, $\frac{a}{2} + \frac{b}{4} \le \sqrt{\frac{27}{8}} = \frac{\sqrt{27}}{\sqrt{8}} = \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{3\sqrt{3} \cdot \sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{3\sqrt{6}}{4}$.
The maximum value is $\frac{3\sqrt{6}}{4}$.
Since $\sqrt{6} \approx 2.449$, the maximum value is approximately $\frac{3 \times 2.449}{4} = \frac{7.347}{4} \approx 1.83675$.
This is not 3. Thus, option (C) is incorrect.

Part (D): To get maximum

This option is incomplete and does not provide a statement or a numerical value, so it cannot be evaluated as correct.

Based on the analysis, options (A) and (B) are correct.