Multivariable calculus -critical points

1. Jul 29, 2006

ultima9999

A rectangular box is to have a volume of $$96m^3$$. The 4 sides and the top are to be made of a material of density $$1kg/m^2$$ and the flat base is to have a density of $$2kg/m^2$$. Suppose that the base is of size x meters by y meters.

It was found in the first question that the mass, M in kg of the box is:
$$M(x,y) = 3xy + \frac{192}{x} + \frac{192}{y}$$

Find the dimensions of the box so that the total mass of the box is a minimum.
So, I just find the critical points, which occur at $$f_x = f_y = 0$$
$$f_x = 3y - \frac{192}{x^2} = 0[tex] [tex]y = \frac{64}{x^2}$$

$$f_y = 3x - \frac{192}{y^2} = 0$$
$$x(x^3 - 1) = 0,\ \ x = 0,\ 1$$

So critical points occur at (1, 64) and x = 0. We cannot have a dimension of 0, so minimum mass at x = 1, y = 64. The next question asks to prove that the dimensions found correspond to a minimum value of M(x,y).

So I use second derivatives and the Hessian matrix.
$$f_{xx} = \frac{384}{x^3},\ \ f_{xy} = 3,\ \ f_{yy} = \frac{384}{y^3}$$
$$\mbox{det(H)} = \frac{384^2}{(xy)^3} - 9$$
At (1, 64), det(H) < 0 which means that it is a saddle point, not a minimum. So, where have I gone wrong, or is the answer simply "there is no minimum"?

Last edited: Jul 29, 2006
2. Jul 29, 2006

benorin

a little help, too much perhaps

Your critical points aren't right: notice how m(x,y)=m(y,x) so that any critical points will have x=y? well, anyhow up to here it is ok:

$$y=\frac{64}{x^2}, x=\frac{64}{y^2}$$​

now assume $$x\neq 0,y\neq 0$$ (because of just what you said: can't have 0 volume), now simplify them to get

$$x^2y=64, xy^2=64$$​

multiply the first equation by y and the second by x to get

$$x^2y^2=64y, x^2y^2=64x$$​

since both equations have one side equal to $$x^2y^2$$ we can set them equal to eachother as in $$64y=64x$$ so x=y, plug this into any previous equation, say $$x^2y=64$$ to get $$x^3=64$$ so x=4, but x=y hence x=y=4 is the critical point.