The radius of the base of a cone is increasing at the rate of 3 cm/minute and the altitude is decreasing at the rate of 4 cm/minute. The rate of change of lateral surface when the radius = 7 cm and altitude 24 cm is

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\text { Let r be the radius, h be the height and S be the lateral surface area of the cone at any time t } . \text { Given }: dr/dt = 3 cm/\min\text { and } dh/dt = - 4 cm/\min \text { Here }, l^2 = h^2 + r^2 \Rightarrow l = \sqrt{\left( 24 \right)^2 + \left( 7 \right)^2} \Rightarrow l = $\sqrt$(625) \Rightarrow l = 25 S=$\pi$ rl \Rightarrow S^2 = \left( $\pi$ rl \right)^2 \Rightarrow S^2 = $\pi$^2 r^2 \left( h^2 + r^2 \right) \Rightarrow S^2 = $\pi$^2 r^4 + $\pi$^2 h^2 r^2 \Rightarrow 2SdS/dt = 4 $\pi$^2 r^3 dr/dt + 2 $\pi$^2 r^2 hdh/dt + 2 $\pi$^2 h^2 rdr/dt \Rightarrow 2$\pi$ rldS/dt = 2 $\pi$^2 rh\left[ 2 r^2/hdr/dt + rdh/dt + hdr/dt \right] \Rightarrow 25dS/dt = 24$\pi$\left[ \frac{2 \left( 7 \right)^2}{24} $\times$ 3 - 7 $\times$ 4 + 24 $\times$ 3 \right] \left[ \text { Given }: r = 7, h = 24 \right] \Rightarrow 25dS/dt = 24$\pi$\left[ $\frac{49}{4}$ - 28 + 72 \right] \Rightarrow 25dS/dt = 24$\pi$\left[ 49 + 288 - $\frac{112}{4}$ \right] \Rightarrow dS/dt = 24$\pi$\left[ $\frac{225}{100}$ \right] \Rightarrow dS/dt = 24$\pi$\left( 2 . 25 \right) \Rightarrow dS/dt = 54$\pi$ \text{cm}^2 /\sec [Solution diagram: 44291_solution_1.png]

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