Question

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse StartFraction x squared Over 25 EndFraction plusStartFraction y squared Over 16 EndFraction equals1 with sides parallel to the coordinate axes.

Answer 1

Answer with Step-by-step explanation:

Let  a rectangle box whose dimensions are u and v.Then, $(u/2, v/2)$

must lie on the ellipse

Given equation of ellipse

$\frac{x^2}{25}+\frac{y^2}{16}=1$

$(u/2,v/2)$must lie on the circle therefore,

$\frac{u^2}{100}+\frac{v^2}{64}=1$

with $u,v\geq 0$

Suppose , we have  to maximize a  function $f(u,v)=uv$ subject to constraints g(u,v)=$\frac{u^2}{100}+\frac{v^2}{64}=1$

$\nabla f=<v,u> \nabla g=<\frac{u}{50},\frac{v}{32} >$

Using Lagrange multipliers method

$\nabla f=\lambda\nabla g$

$v=\lambda \frac{u}{50}$

$u=\lambda\frac{v}{32}$

$\lambda=\frac{50v}{u}=\frac{32u}{v}$

$50v^2=32u^2$

if u=0 then v=0 g(u,v)=$0\neq 1$

It is absurd condition.

Therefore, we take u and v >0

$v^2=\frac{32u^2}{50}=\frac{16}{25}u^2$

Substitute the value in ellipse equation then we get

$\frac{u^2}{100}+\frac{u^2}{100}=1$

$\frac{2u^2}{100}=1$

$u^2=50$

$u=5 \sqrt2$

$v^2=\frac{16}{25}\times 50=32$

$v=4\sqrt2$

The critical point is ($5\sqrt2, 4\sqrt2)$.

Therefore, we concluded that the dimensions of the rectangle of greatest area is attained by choosing a box of dimensions $5\sqrt2 \times 4\sqrt2$.

Answer 2

Answer:

5sqrt(2)

4sqrt(2)

Step-by-step explanation: