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Concept Version 8
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Optimization in Several Variables

To solve an optimization problem, formulate the function f(x,y,⋯)f(x,y, \cdots )f(x,y,⋯) to be optimized and find all critical points first.

Learning Objective

  • Solve a simple problem that requires optimization of several variables


Key Points

    • Mathematical optimization is the selection of a best element (with regard to some criteria) from some set of available alternatives.
    • An optimization process that involves only a single variable is rather straightforward. After finding out the function f(x)f(x)f(x) to be optimized, local maxima or minima at critical points can easily be found. End points may have maximum/minimum values as well.
    • For a rectangular cuboid shape, given the fixed volume, a cube is the geometric shape that minimizes the surface area.

Terms

  • optimization

    the design and operation of a system or process to make it as good as possible in some defined sense

  • cuboid

    a parallelepiped having six rectangular faces


Full Text

Mathematical optimization is the selection of a best element (with regard to some criteria) from some set of available alternatives. An optimization process that involves only a single variable is rather straightforward. After finding out the function f(x)f(x)f(x) to be optimized, local maxima or minima at critical points can be easily found. (Of course, end points may have maximum/minimum values as well.) The same strategy applies for optimization with several variables. In this atom, we will solve a simple example to see how optimization involving several variables can be achieved.

Cardboard Box with a Fixed Volume

A packaging company needs cardboard boxes in rectangular cuboid shape with a given volume of 1000 cubic centimeters and would like to minimize the material cost for the boxes. What should be the dimensions xxx, yyy, zzz of a box?

First of all, the material cost would be proportional to the surface area SSS of the cuboid . Therefore, the goal of the optimization is to minimize a function S(x,y,z)=2(xy+yz+zx)S(x,y,z) = 2(xy + yz+zx)S(x,y,z)=2(xy+yz+zx). The constraint in the case is that the volume is fixed: V=xyz=1000V = xyz = 1000V=xyz=1000.

Rectangular Cuboid

Mathematical optimization can be used to solve problems that involve finding the right size of a volume such as a cuboid.

We will first remove zzz from S(x,y,z)S(x,y,z)S(x,y,z). We can do this by using the constraint z=1000xyz = \frac{1000}{xy}z=​xy​​1000​​. Inserting the expression for zzz in S(x,y,z)S(x,y,z)S(x,y,z) yields:

S(x,y,z)=2(xy+1000x+1000y)\displaystyle{S(x,y,z) = 2\left(xy + \frac{1000}{x} + \frac{1000}{y}\right)}S(x,y,z)=2(xy+​x​​1000​​+​y​​1000​​) 

To find the critical points:

$\displaystyle{\frac{\partial S}{\partial x} = 2 \left(y - \frac{1000}{x^2} \right) = 0\\ \therefore y = \frac{1000}{x^2}}$

and

$\displaystyle{\frac{\partial S}{\partial y} = 2\left(x - \frac{1000}{y^2}\right) = 0\\ \therefore x = \frac{1000}{y^2}}$ 

Then, substituting in the expression found equal to yyy above yields:

 x3=1000x^3 = 1000x​3​​=1000

Therefore, we find that:

 x=y=z=10x=y=z=10x=y=z=10

That is to say, the box that minimizes the cost of materials while maintaining the desired volume should be a 10-by-10-by-10 cube.

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