Question

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid 4x^2+y^2+4z^2=4

a. The length of the box in the x-direction is:_______
b. The length of the box in the y-direction is:_______
c. The length of the box in the z-direction is:_______

Answer 1

Solution :

Let x, y, z be the dimensions of the rectangle.

Volume of the rectangle (V) = xyz

Given that the vertex should lie on ellipse $4x^2 + y^2+4z^2=4$ .......(i)

So here the volume xyz must be maximum with constraints above.

We solve this using Lagranches method with variable λ

Lagranches function is

$F : xyz + \lambda(4x^2+y^2+4z^2-4)=0$ .....(ii)

To find λ, $\frac{dF}{dx}=0; \frac{dF}{dy}=0;\frac{dF}{dz}=0$

$\frac{dF}{dx}=0 \Rightarrow yz+\lambda(16x)=0 \Rightarrow \lambda = -\frac{yz}{16x}$ .............(iii)

$\frac{dF}{dy}=0 \Rightarrow xz+\lambda(2y)=0 \Rightarrow \lambda = -\frac{xz}{2y}$ ..................(iv)

$\frac{dF}{dz}=0 \Rightarrow xy+\lambda(16z)=0 \Rightarrow \lambda = -\frac{xy}{16z}$ ...............(v)

Equating (iii) and (iv)

$-\frac{yz}{16x}=-\frac{xz}{2y} \Rightarrow y^2=8x^2 \Rightarrow y = \sqrt8 x$ ...............(vi)

Equating (iii) and (v)

$-\frac{yz}{16x}=-\frac{xy}{16z} \Rightarrow z^2=x^2 \Rightarrow z = x$ ....................(vii)

Substitute (vi) and (vii) in (i),

From (i),

$4x^2 + (\sqrt8 x)^2+4x^2 = 4$

$\Rightarrow 4x^2 +8x^2+4x^2 = 4 \Rightarrow x = \frac{1}{4}$

From (vi),

$y = \sqrt8 x$

$\Rightarrow y= \sqrt8 \times \frac{1}{4} \Rightarrow y =\frac{\sqrt8}{4}$

$\Rightarrow y =\frac{\sqrt{2\times4}}{4} \Rightarrow y = \frac{1}{\sqrt2}$

From (vii),

z = x

$z =\frac{1}{4}$

Therefore, for maximum volume the dimensions of a rectangle box are

$x =\frac{1}{4} ; y = \frac{1}{\sqrt2} ; z =\frac{1}{4}$