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anzhelika [568]
3 years ago
13

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell

ipse StartFraction x squared Over 25 EndFraction plusStartFraction y squared Over 16 EndFraction equals1 with sides parallel to the coordinate axes.
Mathematics
1 answer:
s2008m [1.1K]3 years ago
8 0

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= \nabla g=

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.

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