Hi, I'm having a hard time with this question.

the stiffness of a beam is proportional to it's width times the cube of it's depth. What are the dimensions of the stiffest beam you an produce from a beam 12" in diameter.

So I draw a rectangle inside a circle, and the diagonal inside the rectangle is 12. The width (w) times the depth cubed equals the stiffness.

S=wd^3

d^2+w^2=144 w=sqr(144-d^2)

S=sqr(144-d^2)(d^3)

S'=sqr(144-d^2)(3d^2)+d^3(1/2(144-d^2)^-1/2)

boils down to -d^2(6d^2+d-864)=0, to find critical points.

d=0, or 6d^2+d-864=0

d=(sqr(20737)-1)/2

w= sqr(144-(sqr(20737)-1729)/12)

did I do this right?

The teacher, tried to get us started on this problem, but following through from where he started, I run into a problem which I don't understand.

The teacher, started by trying to find ds/dw.

Here is where he left us off, S=W(144-w^2)^(3/2)

I found it's derivative, 3w^2sqr(144-w^2)+(144-w^2)^3/2

when I solve for 0, I get w=-12, or 12.

Makes no sense because if the width is 12, the depth is zero. Where did I go wrong.