JEE Challenger
More from Application of Derivatives
MathMedAdv 2015 • Paper 1

Minimize Material Volume of Cylindrical Container with Fixed Capacity

A cylindrical container open at the top is to be fabricated from a solid material subject to the following specifications: it possesses a constant inner volume of V mm3V\text{ mm}^3 and a solid side wall of thickness 2 mm2\text{ mm}. The bottom base is a flat circular disk of thickness 2 mm2\text{ mm} whose radius matches the outer radius of the container.

If the volume of the material utilized to construct the vessel is minimized when the internal radius is 10 mm10\text{ mm}, find the value of V250π\frac{V}{250\pi}.

Official Numerical Answer4

Topics & Concepts

Application of DerivativesMaxima and Minima

Step-by-Step Solution

To find the value of V250π\frac{V}{250\pi}, we determine the volume of the material utilized to construct the cylindrical container as a function of the inner radius rr.

1. Geometric Dimensions and Relations: Let the inner radius of the container be r mmr\text{ mm} and the inner height be h mmh\text{ mm}.

  • The side wall has a thickness of 2 mm2\text{ mm}, so the outer radius is: R=r+2R = r + 2
  • The inner volume of the container is fixed and given by: V=πr2h  ⟹  h=Vπr2V = \pi r^2 h \implies h = \frac{V}{\pi r^2}

2. Volume of the Material: The vessel consists of two parts:

  • The cylindrical side wall: The height of this hollow cylinder is hh, with an outer radius of r+2r + 2 and an inner radius of rr.
Vwall=π[(r+2)2−r2]h=π(4r+4)h=4π(r+1)h\begin{aligned} V_{\text{wall}} &= \pi \left[(r+2)^2 - r^2\right] h \\ &= \pi (4r + 4) h \\ &= 4\pi (r + 1) h \end{aligned}
  • The bottom base: A solid flat circular disk of thickness 2 mm2\text{ mm} and radius equal to the outer radius r+2r+2. Vbase=π(r+2)2×2=2π(r+2)2V_{\text{base}} = \pi (r+2)^2 \times 2 = 2\pi (r+2)^2

Thus, the total volume of the material VmV_m is: Vm(r)=Vwall+Vbase=4π(r+1)h+2π(r+2)2V_m(r) = V_{\text{wall}} + V_{\text{base}} = 4\pi (r+1)h + 2\pi (r+2)^2

Substituting h=Vπr2h = \frac{V}{\pi r^2}: Vm(r)=4π(r+1)(Vπr2)+2π(r+2)2=4V(1r+1r2)+2π(r+2)2V_m(r) = 4\pi (r+1) \left(\frac{V}{\pi r^2}\right) + 2\pi (r+2)^2 = 4V\left(\frac{1}{r} + \frac{1}{r^2}\right) + 2\pi (r+2)^2

3. Minimizing the Material Volume: To find the radius that minimizes the volume of the material, we compute the derivative of Vm(r)V_m(r) with respect to rr and set it to zero: dVmdr=4V(−1r2−2r3)+4π(r+2)\frac{dV_m}{dr} = 4V\left(-\frac{1}{r^2} - \frac{2}{r^3}\right) + 4\pi(r+2)

Given that VmV_m is minimized when the inner radius r=10 mmr = 10\text{ mm}: dVmdr∣r=10=0\left.\frac{dV_m}{dr}\right|_{r=10} = 0

Substitute r=10r = 10: 4V(−1102−2103)+4π(10+2)=04V\left(-\frac{1}{10^2} - \frac{2}{10^3}\right) + 4\pi(10+2) = 0 4V(−101000−21000)+4π(12)=04V\left(-\frac{10}{1000} - \frac{2}{1000}\right) + 4\pi(12) = 0 4V(−121000)+48π=04V\left(-\frac{12}{1000}\right) + 48\pi = 0 −48V1000+48π=0-\frac{48V}{1000} + 48\pi = 0 V1000=π  ⟹  V=1000π\frac{V}{1000} = \pi \implies V = 1000\pi

4. Final Calculation: V250π=1000π250π=4\frac{V}{250\pi} = \frac{1000\pi}{250\pi} = 4

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