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$\begingroup$ @JhoseElijah As the center of the rectangle and sphere is at the origin, if one end of an edge of the rectangular box is is on the sphere at $(x, y, z)$, the other is at $(-x, y, z)$. So the length is $2x$.
"Find the maximum and minimum volumes of a rectangular box whose surface area is $1500 cm^3$ and whose ...
The formula $\rm V=lwh$ means "volume = length times width times height." The variable $\rm l$ is length, the variable $\rm w$ is width, and the variable $\rm h$ is height. Using these, the total area is actually $$\rm 2(l\times h)+2(w\times h)+w^2=1200.$$ We know that $\rm l=w$ (because the base of the box is square), so this is $\rm 4wh+w^2 ...
Carrying this out for all three pairs will lead to $ \ x \ = \ y \ = \ z \ $ : perhaps unsurprisingly, this tells us that the maximal volume box is a cube. $\endgroup$ – colormegone Commented Jun 15, 2014 at 5:20
4. Finding the dimensions of the maximum volume box inside the ellipsoid. I assume that the volume of a box, V(x, y, z) = xyz (they did not give this to me, but this is the volume of a box right?) Ellipsoid: x2 a2 + y2 b2 + z2 c2 = 1. the hint they give me is that. Max volume = 8abc 3√3.
The volume V of the box is V = xyz. So we find max(V) with conditions x + 3y + 6z = 18 and x, y, z ∈ Rnonneg. Use g(x, y, z) = x + 3y + 6z = 18. So. V′(x) = yz = rg′(x) = r. V′(y) = xz = rg′(y) = 3r. V′(z) = xy = rg′(z) = 6r. This means xz = 3yz, xy = 6yz ⇒ z(x − 3y) = 0, y(x − 6z) = 0. If z = 0 or y = 0, then V = 0 and is ...
A box with a square base and an open top is to be made. You have 1200cm2 1200 cm 2 of material to make it. What is the maximum volume the box could have? Here's what I did: 1200 =x2 + 4xz; 1200 = x 2 + 4 x z; where x x is length of base and z z is height of box. Also, let the volume of box be V V, then. V =x2z =x2(1200 −x2 4x) = 300x − (0. ...
4. I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid. x2 4 + y2 9 + z2 16 = 1 x 2 4 + y 2 9 + z 2 16 = 1. A hint given was: If vertex of box in first octants is (x,y,z) then volume is 8xyz. So I first find the partial derivatives.
An open rectangular box having a volume of 108 $\text{cm}^3$ is to be constructed from metal sheets. Find the dimensions of such a box if the amount of metal used in its construction is to be minimised.
$\begingroup$ is there any method to check weather x,y,z yield maximum volume. or we will have to find different ordered pair $\endgroup$ – Naruto_007 Commented Aug 5, 2017 at 10:58