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anonymous
 5 years ago
Multivariable optimiziation?
I have an equation that is rather unwieldy, but can be simplified greatly to k=2n*x*y/(x+y) through substitution of its parts. I have calculated the partial derivatives of k with respect to n, x, and y, and they are the same as the partial derivatives of the original equation (with the exception of a constant factor of 1/100 for the partial derivative of n), and plugging the substitutions back into the equations has given me a set of related rates. I can tell how quickly k increases by increasing any of the other three variables at any one point.
The question I h
anonymous
 5 years ago
Multivariable optimiziation? I have an equation that is rather unwieldy, but can be simplified greatly to k=2n*x*y/(x+y) through substitution of its parts. I have calculated the partial derivatives of k with respect to n, x, and y, and they are the same as the partial derivatives of the original equation (with the exception of a constant factor of 1/100 for the partial derivative of n), and plugging the substitutions back into the equations has given me a set of related rates. I can tell how quickly k increases by increasing any of the other three variables at any one point. The question I h

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, size limits. I'll post the rest of it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The question I have is, if I wanted to substitute the costs of each unit (to be precise, each unit of n has a cost of $2.6, each unit of x has a cost of $15.6, and each unit of y has a cost of $15.4) to calculate the most costefficient way of increasing k at any given moment, how would I do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Original Equation: k=(N/50)(X+100)(Y+100)/(X+Y+200) Substitute n=N/100 x=X+100 y=Y+100 New equation: k=2nxy/(x+y) which can also be written as the harmonic mean m of x and y Mean equation: k=n*m

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dk/dn = (2x*y)/(x+y), but this is off by a factor of 1/100, so dk/dn = (x*y)/(50(x+y)) dk/dx = (2n*y^2)/(x+y)^2 dk/dy = (2n*x^2)/(x+y)^2 Substitute for original values of x, y, and n dk/dn = ((X+100)(Y+100))/(50(X+Y+200)) dk/dx = (2*(N/100)*(Y+100)^2)/(X+Y+200)^2 dk/dy = (2*(N/100)*(X+100)^2)/(X+Y+200)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cost of one unit of N = $2.60 Cost of one unit of X = $15.60 Cost of one unit of Y = $15.40 I understand that each partial derivative is a rate of change for k, but if you just look at the output it almost always appears to be most efficient to purchase X or Y, but this is not the case due to their higher costs. How do I use the costs of X, Y, and N to find out which is most efficient to purchase at any given moment to increase k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have also considered making the cost of X and Y identical at $15.50 for simplicity. I am unsure how greatly it will impact the results if I leave them as they are.
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