A beam of rectangular cross section is to be cut from a log 1.00 m
in diameter. The stiffness of the beam varies directly as the width
and the cube of the depth. What dimensions will give the beam
maximum stiffness?
If the beam has dimensions x and y, then x^2+y^2 = 1
so the stiffness is
s = kx(1-x^2)^(3/2)
now find where ds/dx = 0
To find the dimensions that will give the beam maximum stiffness, we need to maximize the value of the stiffness function. Let's denote the width of the beam as 'w' and the depth as 'd'.
Given that the stiffness varies directly as the width and the cube of the depth, we can express the stiffness function as:
Stiffness = k * w * d^3
where 'k' is a constant of proportionality.
To maximize the stiffness, we can use derivatives. We need to find the critical points of the stiffness function with respect to the variables 'w' and 'd'.
1. Take the partial derivative of the stiffness function with respect to 'w':
d(Stiffness) / dw = k * d^3
2. Set the partial derivative equal to zero and solve for 'd':
0 = k * d^3
Since 'k' is a constant of proportionality and cannot be zero, we can conclude that d = 0.
However, for a beam, the depth cannot be zero, so this solution is not valid.
3. Take the partial derivative of the stiffness function with respect to 'd':
d(Stiffness) / dd = 3k * w * d^2
4. Set the partial derivative equal to zero and solve for 'w':
0 = 3k * w * d^2
w = 0
Again, this solution is not valid as a beam cannot have zero width.
Therefore, after evaluating the partial derivatives, we can conclude that there is no maximum stiffness for this beam. Depending on the constraints of the problem, you may need to consider additional factors or constraints to determine the dimensions that will optimize the stiffness.