DIVIDED in two parts such that first part = x \[\therefore \] Second part = 20-x Now, assume that \[P={{x}^{3}}(20-x)=20{{x}^{3}}-{{x}^{4}}\] Now, \[\frac{dP}{dx}=60{{x}^{2}}-4{{x}^{3}};\] and \[\frac{{{d}^{2}}P}{d{{x}^{2}}}=120x-12{{x}^{2}}\] PUT \[\frac{dP}{dx}=0\] for MAXIMA or minima \[\Rightarrow \frac{dP}{dx}=0\] \[\Rightarrow \,\,4{{x}^{2}}(15-x)=0\Rightarrow x=0,x=15\] \[\therefore {{\left( \frac{{{d}^{2}}P}{d{{x}^{2}}} \right)}_{x=15}}=120\times 15-12\times (225)\] \[=1800-2700=-900<0\] \[\therefore \] P is a maximum at \[x=15\]. \[\therefore \] First part = 15 and second part \[=20-15=15\] Required product \[=15\times 5=75\]