1. The volume of a cube is increasing at a rate of 1200 cm/min at the moment when the lengths of the sides are 20cm. How fast are the lengths of the sides increasing at that [10] moment?

Answer:[tex]1\,\,cm/min[/tex]Step-by-step explanation:Let V be the volume of cube and x be it's side .We know that volume of cube is [tex]\left ( side \right )^{3}[/tex] i.e., [tex]x^3[/tex]Given :[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=1200\,\,cm^3/min[/tex][tex]x=20\,\,cm[/tex]To find : [tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]Solution :Consider equation [tex]V=x^3[/tex]On differentiating both sides with respect to t , we get[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=3x^2\left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )\\1200=3(20)^2\left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )\\\frac{\mathrm{d} x}{\mathrm{d} t} =\frac{1200}{3(20)^2}=\frac{1200}{3\times 400}=\frac{1200}{1200}=1\,\,cm/min[/tex]So,Length of the side is increasing at the rate of [tex]1\,\,cm/min[/tex]