Example of using derivatives to optimize the material cost

by Kurious Fox

Optimization of Material Usage: Engineers use derivatives to minimize the cost of materials while maintaining structural integrity. For example, determining the optimal dimensions of a container to minimize surface area for a given volume.

Example: For a cylindrical container with a fixed volume V = \pi r^2 h , the surface area S = 2\pi r h + 2\pi r^2 . Using the volume constraint, express h in terms of r , differentiate S with respect to r , we can find the minimum value. So,

Given:

  • Volume V = \pi r^2 h
  • Surface area S = 2\pi r h + 2\pi r^2

Solution:

  1. Express h in terms of r using the volume constraint:
    V = \pi r^2 h \implies h = \frac{V}{\pi r^2}
  2. Substitute h into the surface area formula:
    S = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2
  3. Find the first derivative S'(r) to identify critical points:
    S'(r) = \frac{d}{dr} \left(\frac{2V}{r} + 2\pi r^2\right)
    S'(r) = -\frac{2V}{r^2} + 4\pi r
  4. Set the first derivative equal to zero to find critical points:
    -\frac{2V}{r^2} + 4\pi r = 0
    4\pi r = \frac{2V}{r^2}
    4\pi r^3 = 2V
    r^3 = \frac{V}{2\pi}
    r = \left(\frac{V}{2\pi}\right)^{1/3}
  5. Find h using the volume constraint:
    h = \frac{V}{\pi r^2} = \frac{V}{\pi \left(\left(\frac{V}{2\pi}\right)^{2/3}\right)}
    h = \frac{V}{\pi} \cdot \frac{1}{\left(\frac{V}{2\pi}\right)^{2/3}}
    h = \frac{V}{\pi} \cdot \left(\frac{2\pi}{V}\right)^{2/3}
    h = \left(\frac{2V}{\pi}\right)^{1/3}
  6. Verify if this point minimizes the surface area using the second derivative test:
    S''(r) = \frac{d^2}{dr^2} \left(\frac{2V}{r} + 2\pi r^2\right)
    S''(r) = \frac{d}{dr} \left(-\frac{2V}{r^2} + 4\pi r\right)
    S''(r) = \frac{4V}{r^3} + 4\pi
  7. Evaluate S''(r) at r = \left(\frac{V}{2\pi}\right)^{1/3} :
    S''\left(\left(\frac{V}{2\pi}\right)^{1/3}\right) = \frac{4V}{\left(\left(\frac{V}{2\pi}\right)^{1/3}\right)^3} + 4\pi
    S''\left(\left(\frac{V}{2\pi}\right)^{1/3}\right) = \frac{4V}{\frac{V}{2\pi}} + 4\pi
    S''\left(\left(\frac{V}{2\pi}\right)^{1/3}\right) = \frac{4V \cdot 2\pi}{V} + 4\pi
    S''\left(\left(\frac{V}{2\pi}\right)^{1/3}\right) = 8\pi + 4\pi = 12\pi

Since S''(r) = 12\pi > 0 , the critical point r = \left(\frac{V}{2\pi}\right)^{1/3} is indeed a minimum.

Conclusion:

  • The optimal radius r is:
    r = \left(\frac{V}{2\pi}\right)^{1/3}
  • The optimal height h is:
    h = \left(\frac{2V}{\pi}\right)^{1/3}
    These dimensions minimize the surface area of the cylindrical container for a given volume V .

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