What is the largest rectangle that can be inscribed in the first quadrant of the ellipse
I let x and y equal 0 to find the intercepts, but that does not lead to the solution.
1. Draw a sketch.
2. Let l denote the length of the rectangle and w the width. Then the area of the rectangle is:
$\displaystyle a = l \cdot w$
3. Plug in l instead of x and calculate the y-value which is the width of the rectangle:
$\displaystyle w = \frac34 \sqrt{16-l^2}$
4. You now got the area as a function in l:
$\displaystyle a(l)= l \cdot \frac34 \sqrt{16-l^2} = \frac34 \cdot l \cdot \sqrt{16-l^2}$
5. Differentiate a wrt l and solve the equation a'(l) = 0 for l. I've got $\displaystyle l = 2 \sqrt{2}$
6. Now calculate the maxium area.
You can always go with Lagrange multipliers. The equation of constraint is $\displaystyle g=9x^2+16y^2-144$
The area is $\displaystyle A=xy$
You have $\displaystyle \nabla A=\lambda\nabla g$
$\displaystyle \left(y,x\right)=\lambda\left(18x,32y\right)$
$\displaystyle y=18\lambda x$
$\displaystyle x=32\lambda y=32\lambda\times 18\lambda x=24^2\lambda^2x$
$\displaystyle \lambda=\pm\dfrac{1}{24}$
$\displaystyle y=\pm\dfrac{3}{4}x$
$\displaystyle g=9x^2+16\dfrac{9}{16}x^2-144=0$
$\displaystyle 18x^2=144$
$\displaystyle x=\pm \sqrt{8}$ but $\displaystyle x>0$ so $\displaystyle x=\sqrt{8}$
$\displaystyle y=\pm \dfrac{3}{4}x=\pm\dfrac{3}{4}\times \sqrt{8}=\dfrac{3}{\sqrt{2}}$