Your neighbor is growing a sightly different watermelon it also has a rind whose thickness is one tenth of the radius of that watermelon. However, the rind of your neighbors water metons grows at a constant rate of 20 cubic centimeters a week The radus of your neightor's watermelon after 5 weeks is \( \square \) and at that time if is growing at \( \square \)

To solve the problem, we need to determine the radius of your neighbor's watermelon after 5 weeks and the rate at which it is growing at that time. 1. **Understanding the growth of the watermelon:** - The rind thickness is one-tenth of the radius of the watermelon. - The volume of the rind is increasing at a constant rate of 20 cubic centimeters per week. 2. **Volume of the watermelon:** The volume \( V \) of a sphere (the watermelon) is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the watermelon. 3. **Volume of the rind:** The thickness of the rind is \( \frac{1}{10} r \). Therefore, the outer radius of the watermelon including the rind is: \[ R = r + \frac{1}{10} r = \frac{11}{10} r \] The volume of the watermelon including the rind is: \[ V_{\text{total}} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(\frac{11}{10} r\right)^3 = \frac{4}{3} \pi \left(\frac{1331}{1000} r^3\right) = \frac{5324.4}{3000} \pi r^3 \] 4. **Volume of the rind:** The volume of just the rind is the total volume minus the volume of the watermelon: \[ V_{\text{rind}} = V_{\text{total}} - V = \frac{5324.4}{3000} \pi r^3 - \frac{4}{3} \pi r^3 \] To combine these, we need a common denominator: \[ V_{\text{rind}} = \left(\frac{5324.4}{3000} - \frac{4000}{3000}\right) \pi r^3 = \frac{1324.4}{3000} \pi r^3 \] 5. **Rate of change of the volume of the rind:** We know that the volume of the rind is increasing at a rate of 20 cm³ per week: \[ \frac{dV_{\text{rind}}}{dt} = 20 \] Using the chain rule: \[ \frac{dV_{\text{rind}}}{dt} = \frac{dV_{\text{rind}}}{dr} \cdot \frac{dr}{dt} \] We need to find \( \frac{dV_{\text{rind}}}{dr} \): \[ \frac{dV_{\text{rind}}}{dr} = \frac{d}{dr}\left(\frac{1324.4}{3000} \pi r^3\right) = \frac{1324.4}{1000} \pi r^2 \] 6. **Setting up the equation:** Now we can set up the equation: \[ 20 = \frac{1324.4}{1000} \pi r^2 \cdot \frac{dr}{dt} \] Solving for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{20 \cdot 1000}{1324.4 \pi r^2} \] 7. **Finding the radius after 5 weeks:** Since the volume of the rind grows at a constant rate, we can find the total volume of the rind after 5 weeks: \[ V_{\text{rind}}(5) = 20 \cdot 5 = 100 \text{ cm}^3 \] We can set this equal to the volume of the rind: \[ \frac{1324.4}{3000} \pi r^3 = 100 \] Solving for \( r \): \[ r^3 = \frac{100 \cdot 3000}{1324.4 \pi} \] \[ r^3 \approx \frac{300000}{4157.52} \approx 72.1 \] \[ r \approx \sqrt[3]{72.1} \approx 4.16 \text{ cm} \] 8. **Finding the rate of growth at that time:** Substitute \( r \approx 4.16 \) cm into the equation for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{20 \cdot 1000}{1324.4 \pi (4.16)^2} \] Calculate \( (4.16)^2 \approx 17.3056 \): \[ \frac{dr}{dt} \approx \frac{20000}{1324.4 \cdot \pi \cdot 17.3056} \approx \frac{20000}{72267.4} \approx 0.276 \text{ cm/week} \] Thus, after 5 weeks, the radius of your neighbor's watermelon is approximately **4.16 cm**, and at that time, it is growing at a rate of approximately **0.276 cm/week**.