Max Volume of Open Top Box with 300m sq Metal: Solving the Optimization Problem

[Kuma]

Homework Statement

what is the maximum volume of an open top box that can be created with 300m sq of metal assuming none is wasted?

Homework Equations

The Attempt at a Solution

so that means the surface area of the box must total 300 m sq

so

A(l,w,h)= lw + 2lh + 2wh = 300

and volume is given by

V(l,w,h) = lwh

my problem is when taking partials for l w and h in the area equation, I get 0 as the critical points. Are my equations wrong?

 

[Mark44]

Homework Statement

what is the maximum volume of an open top box that can be created with 300m sq of metal assuming none is wasted?

Homework Equations

The Attempt at a Solution

so that means the surface area of the box must total 300 m sq

so

A(l,w,h)= lw + 2lh + 2wh = 300

and volume is given by

V(l,w,h) = lwh

my problem is when taking partials for l w and h in the area equation, I get 0 as the critical points. Are my equations wrong?

Your surface area equation can be solved for one of the variables. You can then substitute for that variable in your volume equation so that volume is a function of only two variables. Take partials of the volume equation, not the surface area equation.

 

[Ray Vickson]

Do you know the method of Lagrange multipliers? Anybody doing optimization should learn this method---it is standard.

RGV

 

[Kuma]

I have done larangian multipliers but only with 2 variables.

anyway I followed the method suggested and got a critical point of

l = 10
h = 5
w = 10

now I need to verify this but I'm unsure of which function to form the hessian with. The volume?

 

[Mark44]

It would have to be the volume. That's what you're finding the maximum value for. The surface area is fixed.

 

[Kuma]

ok I did it for the volume but now I have another problem. The hessian is indicating that its a saddle point rather than a max. This is what I did. The hessian is given by

fll flw flh
fwl fww fwh
fhl fhw fhh

which equals

0 h w
h 0 l
w l 0

which equals

0 5 10
5 0 10
10 10 0

that hessian matrix indicates a saddle. The eigenvalues are not all negative.

did I do something wrong?

 

[Ray Vickson]

ok I did it for the volume but now I have another problem. The hessian is indicating that its a saddle point rather than a max. This is what I did. The hessian is given by

fll flw flh
fwl fww fwh
fhl fhw fhh

which equals

0 h w
h 0 l
w l 0

which equals

0 5 10
5 0 10
10 10 0

that hessian matrix indicates a saddle. The eigenvalues are not all negative.

did I do something wrong?

You are using the wrong second-order test. You must either (1) use an unconstrained method (for example, minimizing F(l,w), where F = the value of V when h is eliminated by using the constraint to solve for h as a function of l and w---as suggested by Mark44; or(2) use the Lagrange multiplier method and keep all three variables l, w and h. In case (1) your function F(l,w) is being minimized using unconstrained criteria, so the Hessian of that 2-dimensional function should be used. That is not what you did. In case (2) the appropriate Hessian to test is the Hessian of the LAGRANGIAN, not the objective function. (By objective function we mean: the thing you are trying to maximize or minimize.) Furthermore, we test for positive or negative definiteness of this Hessian projected down into the tangent subspace of the constraint, not in the whole space. You did not do that, either. (When I did it, I got a negative-definite projected Hessian of the Lagrangian, so the point I found---the same as yours---is a strict local constrained maximum, by some appropriate theorems in Optimization theory.) So, unless you want to deal with projected Hessians, you had better use the lower-dimensional unconstrained version.

RGV

 

[AndreTheGiant]

I have a similar problem, so then in this one the hessian should be used for the Volume function in terms of l and w or 2 variables instead of all 3?