Optimization: Minimum Surface Area

In summary, the conversation discusses an optimization problem involving a hexagonal prism and a trihedral pyramid. The surface area of the object must be minimized for a given volume, with variables including A, a, h, and x. The equations for surface area and volume are provided, leading to a final simplified equation in terms of x. The conversation also discusses using calculus to find the displacement x that yields the minimum surface area.
  • #1
JoeyC2488
4
0

Homework Statement



I need help on an optimization problem involving a hexagonal prism with no bottom or top, but the top is covered by a trihedral pyramid which has a displacement, x, such that the surface area of the object is at a minimum for a given volume. The assigned variables include:
A= the surface area of the object
a= the length of each side of the hexagon (a=1)
h= the height of the prism
x= displacement

Homework Equations



Given that A= 6(ah-1/2ax) + 3a[tex]\sqrt{3}[/tex]*[tex]\sqrt{x^2+(a^2/4)}[/tex], use calculus to find the displacement, x that yields the minimum surface area. Calculate x if a=1.


The Attempt at a Solution



I'm struggling trying to find a secondary equation for this optimization problem and I think I need to use an equation evolving h, but I'm not sure what.
 
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  • #2
The volume V is to be constant, this will lead you to an equation in h and x, from which you may find an expression for h.
 
  • #3
How would I use a formula for volume since the problem works with two shapes.

V=(3[tex]\sqrt{3}[/tex])/2 a^2*h

h= V/ (3[tex]\sqrt{3}[/tex]/2 a^2*h

Then, I can just plug this into the first equation, simplify, and start the calculus. Is that right?
 
  • #4
Your TOTAL volume is the sum of the prism's volume and the pyramid's volume.

That is the equation in h and x you are to set up!
 
  • #5
Your total volum is therefore:
[tex]V=\frac{3\sqrt{3}}{2}a^{2}h+\frac{1}{3}\frac{3\sqrt{3}}{2}a^{2}x\to{h}=\frac{2V}{3\sqrt{3}a^{2}}-\frac{x}{3}[/tex]
 
  • #6
So after substituting that in for h, the simplified form would be:

A= 4V-5x[tex]\sqrt{3}[/tex]+ 9[tex]\sqrt{x^2+4}[/tex]

[tex]\sqrt{3}[/tex]
 
  • #7
How would I take the derivative of this equation then with multiple variables?
 
  • #8
You only have one variable, x. Remember, V is a constant!
 

FAQ: Optimization: Minimum Surface Area

1. What is optimization and why is it important?

Optimization is the process of finding the most efficient solution to a given problem. It is important because it helps us make the best use of resources and achieve the desired outcome with minimum effort or cost.

2. How is minimum surface area optimization used in real life?

Minimum surface area optimization is used in various real-life applications such as designing efficient packaging, minimizing material usage in construction, and creating streamlined shapes for vehicles to reduce drag and increase fuel efficiency.

3. What are the key factors to consider in minimum surface area optimization?

The key factors to consider in minimum surface area optimization include the shape, size, and material of the object being optimized, as well as any constraints or limitations that may affect the optimization process.

4. What are the common methods for solving minimum surface area optimization problems?

Common methods for solving minimum surface area optimization problems include calculus-based methods, such as the method of Lagrange multipliers, and computer-based optimization algorithms, such as genetic algorithms and simulated annealing.

5. Can optimization lead to a perfect solution?

No, optimization can lead to a near-perfect solution, but it cannot guarantee a perfect solution. This is because there may be multiple optimal solutions or the problem may be too complex to be solved exactly.

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