# Thread: Optimization [Derivatives]

1. ## Optimization [Derivatives]

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.

2. Originally Posted by castle
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:

$a = l \cdot w$

3. Plug in l instead of x and calculate the y-value which is the width of the rectangle:

$w = \frac34 \sqrt{16-l^2}$

4. You now got the area as a function in l:

$a(l)= l \cdot \frac34 \sqrt{16-l^2} = \frac34 \cdot l \cdot \sqrt{16-l^2}$

5. Differentiate a wrt l and solve the equation a'(l) = 0 for l. I've got $l = 2 \sqrt{2}$

6. Now calculate the maxium area.

3. You can always go with Lagrange multipliers. The equation of constraint is $g=9x^2+16y^2-144$

The area is $A=xy$

You have $\nabla A=\lambda\nabla g$

$\left(y,x\right)=\lambda\left(18x,32y\right)$

$y=18\lambda x$

$x=32\lambda y=32\lambda\times 18\lambda x=24^2\lambda^2x$

$\lambda=\pm\dfrac{1}{24}$

$y=\pm\dfrac{3}{4}x$

$g=9x^2+16\dfrac{9}{16}x^2-144=0$

$18x^2=144$

$x=\pm \sqrt{8}$ but $x>0$ so $x=\sqrt{8}$

$y=\pm \dfrac{3}{4}x=\pm\dfrac{3}{4}\times \sqrt{8}=\dfrac{3}{\sqrt{2}}$

4. Originally Posted by earboth

$w = \frac34 \sqrt{16-l^2}$

I know it is a silly question, but can you show me step by step how you got that. Thank You.

5. Originally Posted by castle
I know it is a silly question, but can you show me step by step how you got that. Thank You.
Here we go:

Solve for y:

$\begin{array}{rcl}9l^2+16w^2&=&144 \\ 16w^2&=&144-9l^2 \\ w^2&=&\dfrac{144-9l^2}{16} \\ w^2&=&\dfrac{9(16-l^2)}{16}\end{array}$

... and now calculate the square-root of both sides of the equation.

6. O.K. Thanks.