What is the product of two parts of 20, such that the product of one part and the cube of the other is maximum ?

Solution :
Let 20 be divided in two parts such that first part $$ = x$$
$$\therefore $$  Second part $$ = 20 – x$$
Now, assume that $$P = {x^3}\left( {20 - x} \right) = 20{x^3} - {x^4}$$
Now, $$\frac{{dP}}{{dx}} = 60{x^2} - 4{x^3}\,;\,{\text{and }}\frac{{{d^2}P}}{{d{x^2}}} = 120x - 12{x^2}$$
Put $$\frac{{dP}}{{dx}} = 0$$   for maxima or minima
$$\eqalign{ & \Rightarrow \frac{{dP}}{{dx}} = 0 \cr & \Rightarrow 4{x^2}\left( {15 - x} \right) = 0 \cr & \Rightarrow x = 0,\,x = 15 \cr & \therefore \,{\left( {\frac{{{d^2}P}}{{d{x^2}}}} \right)_{x = 15}} = 120 \times 15 - 12 \times \left( {225} \right) \cr & = 1800 - 2700 \cr & = - 900 < 0 \cr} $$
$$\therefore \,P$$  is a maximum at $$x = 15.$$
$$\therefore $$  First part $$ = 15$$  and second part $$ = 20 – 15 = 5$$
Required product $$ = 15 \times 5 = 75$$