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Question Number 218813 by Spillover last updated on 15/Apr/25

Answered by nikif99 last updated on 15/Apr/25

Commented by Spillover last updated on 16/Apr/25

$$greatwork.thanks$$

Answered by mr W last updated on 16/Apr/25

Commented by mr W last updated on 16/Apr/25

Commented by mr W last updated on 16/Apr/25

$$h_1=R-13$$$$V_W=\frac{2\pi R^3}{3}-\pi h_1^2(R-\frac{h_1}{3})$$$$=\frac{2\pi R^3}{3}-\pi {(R-13)}^2(R-\frac{R-13}{3})$$$$=\frac{2\pi R^3}{3}-\frac{\pi {(R-13)}^2(2R+13)}{3}$$$$V_W={17}^2\pi (R-\frac{17}{3})=\frac{{17}^2\pi (3R-17)}{3}$$$$\frac{2\pi R^3}{3}-\frac{\pi {(R-13)}^2(2R+13)}{3}=\frac{{17}^2\pi (3R-17)}{3}$$$$39R^2-867R+2716=0$$$$\Rightarrow R=\frac{867+\sqrt{{867}^2-4\times 39\times 2716}}{2\times 39}\approx 18.458>17✓$$$$$$$$2V_W=\pi h^2(R-\frac{h}{3})$$$$\pi h^2(R-\frac{h}{3})=2\times {17}^2\pi (R-\frac{17}{3})$$$$h^2(3R-h)=2\times {17}^2(3R-17)$$$$\Rightarrow h\approx 28.9997\approx 29cm$$

Commented by Spillover last updated on 16/Apr/25

$$greatwork.thanks$$

Answered by Spillover last updated on 17/Apr/25

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