Calc Help, I'm not sure the problem has enough info to solve for what its asking...Please Help! :-)?

Question

The strength of a rectangular beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangulare beam that can be cut from a 12 in. diameter log.



I assumed the strongest beam to be a square rectangle, with sides eqaul to the square root of 72...But I'm not sure how to solve for the depth at all, and i dont know if I need to equations to relate to determine the maximum of the function. Please Help.

Answer 1

The depth is like the height of the beam. Relating the width and depth w^2 + d^2 = 144

We want to maximize F = kwd^2. Where k is the constant of proportionality. It is easier to maximize the function F = wd^2

Solve first equation for d^2.
d^2 = 144 - w^2

F = w(144 - w^2)
F = 144w - w^3
F' = 144 -3w^2
0 = 144 - 3w^2
3w^2 = 144
w^2 = 48
w= 4sqrt(3)
w^2 + d^2 = 144
48 + d^2 = 144
d^2 = 96
d = 4sqrt(6)

The beam is not square, because the strength is not proportional to the width times the depth. The depth is squared.

Answer 2

Without getting into the gory details of the solution, the key here is that the beam is cut from the interior of a 12-inch diameter log. The width is one leg, the depth is the other. When the beam is loaded, the width x length is the face on which the beam is loaded, and the depth is the the distance from the top to the bottom of the beam. The two are related as sides of a right triangle whose hypotenuse =12.

Answer 3

Won't be a square ... should be "taller" than it is wide.

Strength s = w*d^2

Now, the "diagonal" will equal the diameter of the log:
w^2 + d^2 = 144

Substituting: w^2 + s/w = 144
s = w*(144-w^2) = 144w -w^3
s' = 144 - 3w^2, so s' = 0 for w=sqrt(144/3)=sqrt(48)
s'' = -6w, so s''<0 for w>0, so w gives maximum s

w = sqrt(48) = 4sqrt(3)
d = sqrt(144-48) = sqrt(96) = 4sqrt(6)

Answer 4

Assumption: The the pass section is a rectangle inscribed in a circle of 12" diameter. permit w be the width and h the top. For an inscribed rectangle, the diagonal = diameter of the circle. capability is proportional to wh^2. we could desire to maximise this. w^2 + h^2 = 12^2, from which h^2 = one hundred forty four - w^2 wh^2 = w x (one hundred forty four - w^2) = 144w - w^3 = s (say) for s to be optimal ds/dw = 0 ds/dw = one hundred forty four -3w^2 = 0 from which w^2 = one hundred forty four/3 = 40 8 w = sqrt 40 8 = 6.ninety 3" h = sqrt (one hundred forty four - w^2) = sqrt (one hundred forty four - 40 8) = sqrt ninety six = 9.8" The beam has a pass section having: width = 6.ninety 3" and intensity = 9.8".

Answer 5

So, S = wd^2
And w^2 + d^2 = 144
so d^2 = 144 - w^2
and S = w(144-w^2) = 144w - w^3

dS/dw = 144 - 3w^2 so S has a maximum when it's 0

144- 3w^2 = 0
144=3w^2
48=w^2 so w = √48 making d be √96