Optimization [Derivatives]

**castle** · July 14th 2010, 08:37 AM

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.

**earboth** · July 14th 2010, 10:02 AM

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.

**fobos3** · July 14th 2010, 10:19 AM

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 $x=\sqrt{8}$

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

**castle** · July 14th 2010, 01:15 PM

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.

**earboth** · July 14th 2010, 09:14 PM

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.

**castle** · July 15th 2010, 06:10 AM

O.K. Thanks.

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