TITLE: Optimization of Rectangular Beam
QUESTION [1 upvotes]: A rectangular beam has breadth $b$ mm and depth $d$ mm. Its strength is proportional to both the square root of its breadth and its depth cubed. What are the dimensions (in mm) of the strongest beam whose cross sectional diagonal is $q$ mm?
A. $d=\sqrt{\frac{3}{7}}q$ , $b = \sqrt{\frac{3}{7}}q$
B. $d=\frac{3}{\sqrt{7}}q$ , $b=\frac{3}{\sqrt{7}}q$
C. $d=\frac{q}{\sqrt{7}}$ , $b=\frac{q}{\sqrt{7}}$
D. $d=\sqrt{\frac{6}{7}}q$ , $b=\frac{q}{\sqrt{7}}$
E. $d=q$ , $b=3q$
What I've attempted so far:
Let the strength be $S$, since $S \varpropto \sqrt{B}$ and $S \varpropto d^3$, $S \varpropto \sqrt{b}d^3$ $\implies$ $S=k\sqrt{B}d^3$.
For q, since beam is rectangular $q$ would be given by the Pythagorean Theorem, that is $q^2 = b^2 + d^2$
Honestly not quite sure where to go from here, I do believe since $S$ is maximum, differentiating $S$ would have something to do with it, where then the derivative = 0. However, what would S be a function of in order to differentiate it?

REPLY [0 votes]: I think, for answer A,B,C,E, the diagonal is not $q$?
You have already give $$S=k \sqrt{b}d^3$$ and $$q^2=b^2+d^2$$ so $$S=k\sqrt{b}(\sqrt{q^2-b^2})^{\frac{3}{2}}$$ with $$\frac{\partial S}{\partial b}=0$$ there are 4 solutions, the only one that both $b>0,d>0$ is $$b=\frac{q}{\sqrt{7}}, d=q\sqrt{\frac{6}{7}}$$ gives D.
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