36. a rectangle is to be inscribed in the ellipse x² + 4y² = 16. what is the maximum possible area of the…

Explanation:

Step1: Rewrite ellipse equation

The ellipse equation $x^{2}+4y^{2}=16$ can be written as $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$, so $a = 4$ and $b = 2$. Let the vertex of the rectangle in the first - quadrant on the ellipse be $(x,y)$. Then the area of the rectangle $A=4xy$.

Step2: Express $y$ in terms of $x$

From $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$, we get $y = \frac{1}{2}\sqrt{16 - x^{2}}$. So $A(x)=2x\sqrt{16 - x^{2}},0\lt x\lt4$.

Step3: Square the function for simplicity

Let $f(x)=A^{2}(x)=4x^{2}(16 - x^{2})=64x^{2}-4x^{4}$.

Step4: Find the derivative

Differentiate $f(x)$ with respect to $x$. $f^\prime(x)=128x - 16x^{3}=16x(8 - x^{2})$.

Step5: Find critical points

Set $f^\prime(x)=0$. Then $16x(8 - x^{2})=0$. Since $x\gt0$, we solve $8 - x^{2}=0$, getting $x = 2\sqrt{2}$.

Step6: Find the value of $y$

When $x = 2\sqrt{2}$, from $y=\frac{1}{2}\sqrt{16 - x^{2}}$, we have $y=\sqrt{2}$.

Step7: Calculate the maximum area

$A = 4xy$, substituting $x = 2\sqrt{2}$ and $y=\sqrt{2}$, we get $A = 16$.

Answer:

16