To solve this problem, we need to minimize the cost of construction given the constraints. The total volume of the silo is the sum of the volume of the cylinder and the hemisphere. The cost is determined by the surface area, with the hemisphere being more expensive. We will express the cost in terms of the radius and height, and use calculus to find the dimensions that minimize this cost.
The volume V V V of the silo, which consists of a cylindrical part and a hemispherical part, is given by the equation: V=πr2h+23πr3 V = \pi r^2 h + \frac{2}{3} \pi r^3 V=πr2h+32πr3 Rearranging this equation to express h h h in terms of r r r and V V V, we have: h=Vπr2−23r h = \frac{V}{\pi r^2} - \frac{2}{3} r h=πr2V−32r
The cost C C C of constructing the silo is determined by the surface area, with the cost of the hemisphere being 10 times that of the cylindrical sidewall. The cost function can be expressed as: C=2πrh+10(2πr2) C = 2\pi r h + 10(2\pi r^2) C=2πrh+10(2πr2) Substituting the expression for h h h into the cost function gives: C=20πr2+2πr(Vπr2−23r) C = 20\pi r^2 + 2\pi r \left(\frac{V}{\pi r^2} - \frac{2}{3} r\right) C=20πr2+2πr(πr2V−32r) This simplifies to: C=20πr2+2Vr−4π3r2 C = 20\pi r^2 + \frac{2V}{r} - \frac{4\pi}{3} r^2 C=20πr2+r2V−34πr2
To minimize the cost, we differentiate C C C with respect to r r r and set the derivative equal to zero: dCdr=2πr(−Vr3−43)+40πr+−2Vr2=0 \frac{dC}{dr} = 2\pi r \left(-\frac{V}{r^3} - \frac{4}{3}\right) + 40\pi r + \frac{-2V}{r^2} = 0 drdC=2πr(−r3V−34)+40πr+r2−2V=0 Solving this equation yields the critical points: r=0.2574V1/3 r = 0.2574 V^{1/3} r=0.2574V1/3 The other critical points are complex and not relevant for this physical problem.
Using the critical point r=0.2574V1/3 r = 0.2574 V^{1/3} r=0.2574V1/3, we can substitute back into the expression for h h h: h=Vπ(0.2574V1/3)2−23(0.2574V1/3) h = \frac{V}{\pi (0.2574 V^{1/3})^2} - \frac{2}{3} (0.2574 V^{1/3}) h=π(0.2574V1/3)2V−32(0.2574V1/3) Calculating this will give us the height corresponding to the optimal radius.
The optimal radius and height for the silo are: r=0.2574V1/3,h=Vπ(0.2574V1/3)2−23(0.2574V1/3) r = 0.2574 V^{1/3}, \quad h = \frac{V}{\pi (0.2574 V^{1/3})^2} - \frac{2}{3} (0.2574 V^{1/3}) r=0.2574V1/3,h=π(0.2574V1/3)2V−32(0.2574V1/3) Thus, the final boxed answers are: r=0.2574V1/3,h=Vπ(0.2574V1/3)2−23(0.2574V1/3) \boxed{r = 0.2574 V^{1/3}, \quad h = \frac{V}{\pi (0.2574 V^{1/3})^2} - \frac{2}{3} (0.2574 V^{1/3})} r=0.2574V1/3,h=π(0.2574V1/3)2V−32(0.2574V1/3)
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