Question

Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid x281+y29+z225=1 Hint: By symmetry, you can restrict your attention to the first octant (where x,y,z≥0), and assume your volume has the form ????=8xyz. Then arguing by symmetry, you need only look for points which achieve the maximum which lie in the first octant. Maximum volume:

Answer 1

Answer:

Detailed step-wise solution is given below for better demonstration:

Answer 2

The volume of the rectangular box is given by using the Lagrangian multiplier theorem is 172.435 cubic units.

What is the Lagrangian multiplier theorem?

It states that any local maxima or any local minima of the function are calculated under the equality constraints.

The equation of ellipsoid is given as

$\rm \dfrac{x^{2} }{4} + \dfrac{y^{2} }{64}+\dfrac{z^{2} }{49} = 1$ ... 1

Let the edges of the required rectangular box be x, y, and z.

Then, the volume of the box in the first quadrant is V = xyz

Then we have

$\rm \phi (x,y,z) = \dfrac{x^{2} }{4} + \dfrac{y^{2} }{64}+\dfrac{z^{2} }{49} -1 = 0$

By the Lagrange multiplier

$\begin{aligned} \bigtriangledown V &= \lambda \bigtriangledown \phi\\\\(V_x,V_y,V_z) &= \lambda (\phi _x,\phi _y, \phi _z)\\\\(yz,xz,xy) &= \lambda (\dfrac{2x}{4},\dfrac{2y}{64},\dfrac{2z}{49}) \\\\\end{aligned}\\yz = \lambda \dfrac{x}{2}, xz = \lambda \dfrac{y}{32}, xy = \lambda \dfrac{2z}{49}\\\\xyz = \lambda \dfrac{x^{2} }{2}, xyz = \lambda \dfrac{y^2}{32}, xyz = \lambda \dfrac{2z^2}{49}$

$\rightarrow \lambda \dfrac{x^2 }{2} = \lambda \dfrac{y^2 }{32} =\lambda \dfrac{2z^2 }{49}\\\\\rightarrow \dfrac{x^2}{2} = \dfrac{y^2}{32} =\dfrac{2z ^2}{49} = k \ \ \ (since \lambda \neq 0)\\\\\rightarrow x^2= 2k , y^2=32k, z^2 = \dfrac{49k}{2}$... 2

Put in equation 1, we have

$\begin{aligned} \dfrac{x^{2} }{4} + \dfrac{y^{2} }{64}+\dfrac{z^{2} }{49} -1 &= 0\\\\\dfrac{2k }{4} + \dfrac{32k }{64}+\dfrac{ 49k}{2*49} -1 &= 0\\\\k (\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}) &= 1\\\\k &= \dfrac{2}{3} \end{aligned}$

Put k = 2/3 in equation 2, we have

$\rm x^2 = 2* \dfrac{2}{3}, y^2 = 32* \dfrac{2}{3}, z^2 = \dfrac{49}{2}* \dfrac{2}{3} \\\\x = \pm \dfrac{2}{\sqrt{3}},y = \pm \dfrac{8}{\sqrt{3}}, z = \pm \dfrac{7}{sqrt{3}}$

Then the volume of the rectangular box will be

$\rm Volume = \dfrac{2}{\sqrt{3}}*\dfrac{8}{\sqrt{3}}*\dfrac{7}{\sqrt{3}}\\\\Volume = \dfrac{896}{3 \sqrt{3}} = 172.435$

More about the Lagrangian multiplier theorem link is given below.

https://brainly.com/question/7227603