(adsbygoogle = window.adsbygoogle || []).push({}); A rectangular box is to have a volume of [tex]96m^3[/tex]. The 4 sides and the top are to be made of a material of density [tex]1kg/m^2[/tex] and the flat base is to have a density of [tex]2kg/m^2[/tex]. 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:

[tex]M(x,y) = 3xy + \frac{192}{x} + \frac{192}{y}[/tex]

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 [tex]f_x = f_y = 0[/tex]

[tex]f_x = 3y - \frac{192}{x^2} = 0[tex]

[tex]y = \frac{64}{x^2}[/tex]

[tex]f_y = 3x - \frac{192}{y^2} = 0[/tex]

[tex]x(x^3 - 1) = 0,\ \ x = 0,\ 1[/tex]

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.

[tex]f_{xx} = \frac{384}{x^3},\ \ f_{xy} = 3,\ \ f_{yy} = \frac{384}{y^3}[/tex]

[tex]\mbox{det(H)} = \frac{384^2}{(xy)^3} - 9[/tex]

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"?

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# Multivariable calculus -critical points

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