Example on max/min functions

The strenght of a wooden beam is given by [math]S=k\cdot width\cdot depth^2[/math], where [math]k>0[/math] depends on the type of wood(it is a constant). What is the strongest beam that can be cut from a log with diameter 12cm?[br][br]The function provided by the exercise has two variables, but to apply our [br]Calculus methods, we need to transform it into a function of a single [br]variable. Knowing the beam will have a rectangular base, we can use [br]Pythagoras' theorem to find [math]w[/math] as a function of [math]d[/math].[br][br][math]12^2=d^2+w^2\rightarrow144=d^2+w^2\rightarrow d^2=w^2-144[/math]. [br]Substituting the value in S we get [math]S(w)=k\cdot w\cdot\left(144-w^2\right)=k\cdot\left(144w-w^3\right)[/math] , and its derivative [math]S'\left(w\right)=k\cdot\left(144-3w^2\right)[/math]
With a single variable function, the next step is to find the maximum values of the function on the interval [math][0,12][/math], i.e. the critical points on that interval. The graph below shows [math]S(w)[/math] in green and its derivative [math]S'(w)[/math] in orange, use it to find the maximum of the function on [math][0,12][/math] (use an approximation if necessary).
Using the value of the critical point, adjust the sliders in the graph below and compare the value of the area found by min/maxing the problem and other values of [math]w[/math] and [math]d[/math].[br][br]Is it possible to find other values for [math]w[/math] and [math]d[/math], on that interval, that creates a stronger beam?

Information: Example on max/min functions