Question

If $x^2 + x = 1 - y^2$, where $x > 0, y > 0$ and maximum value of $x\sqrt{y}$ is $\frac{1}{a\sqrt{a}}$, then value of $a^5$ is

• A. 16
• B. 32
• C. 64
• D. 128

Answer 1

Answer: (B)

32

Answer 2

The constraint $x^2 + x = 1 - y^2$ can be rewritten as $x^2 + x + y^2 = 1$. Completing the square for $x$ terms gives $(x + \frac{1}{2})^2 + y^2 = \frac{5}{4}$. We want to maximize $f(x,y) = x\sqrt{y}$. From the constraint, $y^2 = 1 - x - x^2$, so $y = \sqrt{1 - x - x^2}$ for $y>0$. The expression to maximize becomes $x(1 - x - x^2)^{1/4}$. To simplify, we can maximize its fourth power: $G(x) = x^4(1 - x - x^2) = x^4 - x^5 - x^6$. Taking the derivative and setting it to zero: $G'(x) = 4x^3 - 5x^4 - 6x^5 = x^3(4 - 5x - 6x^2) = 0$. Since $x>0$, we solve $6x^2 + 5x - 4 = 0$. Using the quadratic formula, $x = \frac{-5 \pm \sqrt{25 - 4(6)(-4)}}{12} = \frac{-5 \pm \sqrt{121}}{12} = \frac{-5 \pm 11}{12}$. Since $x>0$, we get $x = \frac{6}{12} = \frac{1}{2}$. Substituting $x=\frac{1}{2}$ into $y^2 = 1 - x - x^2$, we get $y^2 = 1 - \frac{1}{2} - (\frac{1}{2})^2 = 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Since $y>0$, $y=\frac{1}{2}$. The maximum value of $x\sqrt{y}$ is $\frac{1}{2}\sqrt{\frac{1}{2}} = \frac{1}{2\sqrt{2}}$. We are given this equals $\frac{1}{a\sqrt{a}}$. Thus, $a\sqrt{a} = 2\sqrt{2}$, which implies $a^{3/2} = 2^{3/2}$, so $a=2$. We need to find $a^5 = 2^5 = 32$.