Let's break down the given problem step by step: **Given:** - The two radii of the frustum, \( r \) and \( R \), are increasing at a rate of \( 8 \) cm/min. - The height, \( h \), is increasing at \( 18 \) cm/min. - At the instant of interest, \( r = 15 \) cm, \( R = 25 \) cm, \( h = 10 \) cm. We are asked to find: 1. \( \frac{dV}{dt} \) (rate of change of volume) 2. \( \frac{dS}{dt} \) (rate of change of surface area) --- ## 1. **Volume of a Frustum** The volume \( V \) of a frustum of a cone: \[ V = \frac{1}{3} \pi h (R^2 + Rr + r^2) \] Differentiate with respect to \( t \): \[ \frac{dV}{dt} = \frac{1}{3} \pi \left[ (R^2 + Rr + r^2) \frac{dh}{dt} + h \frac{d}{dt}(R^2 + Rr + r^2) \right] \] Now, \[ \frac{d}{dt}(R^2 + Rr + r^2) = 2R \frac{dR}{dt} + (R \frac{dr}{dt} + r \frac{dR}{dt}) + 2r \frac{dr}{dt} \] \[ = 2R \frac{dR}{dt} + R \frac{dr}{dt} + r \frac{dR}{dt} + 2r \frac{dr}{dt} \] Since \( \frac{dr}{dt} = \frac{dR}{dt} = 8 \) cm/min, substitute the values: \[ = 2R(8) + R(8) + r(8) + 2r(8) = 8(2R + R + r + 2r) = 8(3R + 3r) = 24(R + r) \] Now plug in the values: - \( r = 15 \) - \( R = 25 \) - \( h = 10 \) - \( \frac{dh}{dt} = 18 \) So, \[ V = \frac{1}{3}\pi h (R^2 + Rr + r^2) \] \[ R^2 + Rr + r^2 = 25^2 + 25 \times 15 + 15^2 = 625 + 375 + 225 = 1225 \] \[ \frac{dV}{dt} = \frac{1}{3}\pi \left[1225 \times 18 + 10 \times 24 \times (25 + 15) \right] \] \[ = \frac{1}{3}\pi \left[22050 + 10 \times 24 \times 40\right] \] \[ = \frac{1}{3}\pi \left[22050 + 9600\right] \] \[ = \frac{1}{3}\pi (31650) \] \[ = 10550\pi \ \text{cm}^3/\text{min} \] --- ## 2. **Surface Area of a Frustum** The lateral surface area \( S \) of a frustum: \[ S = \pi (R + r) l \] where \( l \) is the slant height: \[ l = \sqrt{(R - r)^2 + h^2} \] Differentiate \( S \) with respect to \( t \): \[ \frac{dS}{dt} = \pi \left[(R + r) \frac{dl}{dt} + l \left(\frac{dR}{dt} + \frac{dr}{dt}\right)\right] \] Now, \[ \frac{dl}{dt} = \frac{1}{2} \left[(R - r)^2 + h^2\right]^{-1/2} \cdot \left[2(R - r)\left(\frac{dR}{dt} - \frac{dr}{dt}\right) + 2h\frac{dh}{dt}\right] \] \[ = \frac{(R - r)(\frac{dR}{dt} - \frac{dr}{dt}) + h\frac{dh}{dt}}{\sqrt{(R - r)^2 + h^2}} \] Given \( \frac{dR}{dt} = \frac{dr}{dt} = 8 \), so \( (\frac{dR}{dt} - \frac{dr}{dt}) = 0 \): \[ \frac{dl}{dt} = \frac{h \frac{dh}{dt}}{l} \] Calculate \( l \): \[ l = \sqrt{(25 - 15)^2 + 10^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \] So, \[ \frac{dl}{dt} = \frac{10 \times 18}{10\sqrt{2}} = \frac{18}{\sqrt{2}} = 9\sqrt{2} \] Now plug in all values: - \( R + r = 40 \) - \( l = 10\sqrt{2} \) - \( \frac{dR}{dt} + \frac{dr}{dt} = 16 \) \[ \frac{dS}{dt} = \pi \left[40 \times 9\sqrt{2} + 10\sqrt{2} \times 16 \right] \] \[ = \pi \left[360\sqrt{2} + 160\sqrt{2}\right] \] \[ = \pi (520\sqrt{2}) \] --- ## **Final Answers:** \[ \boxed{ \begin{aligned} \frac{dV}{dt} &= 10550\pi \ \text{cm}^3/\text{min} \\ \frac{dS}{dt} &= 520\sqrt{2}\pi \ \text{cm}^2/\text{min} \end{aligned} } \] Let me know if you'd like these values in decimal form!

The rates of change of volume and surface area of a frustum are found by differentiating their formulas with respect to time, considering the rates of change of the radii, height, and slant height. These derivatives involve applying the chain rule to account for how each geometric parameter varies over time.