Optimization

engineertobe

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Oct 8, 2011
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The problem is:

Find three positive numbers x, yand z such that the sum of y,z, and 5 times x is equal to 180 and the product of the three numbers is maximum.

I understand the basics of the problem that you have (y+z+5x)=180 but how do you work the problem to maximize the product of the three numbers?
 
Are you familiar with Lagrange multipliers?. You can also use differentiation.

I will use multipliers.

We want to maximize f(x,y,z)=xyz\displaystyle f(x,y,z)=xyz subject to the constraint g(x,y,z)=5x+y+z=180\displaystyle g(x,y,z)=5x+y+z=180

Differentiate:

f=yzi+xzj+xyk=λ(5i+j+k)\displaystyle \nabla f=yzi+xzj+xyk=\lambda(5i+j+k)

Equate:

yz=5λ\displaystyle yz=5\lambda..........[1]

xz=λ\displaystyle xz=\lambda............[2]

xy=λ\displaystyle xy=\lambda............[3]

Set [1] and [2] equal and we find

yz5=xzx=y5\displaystyle \frac{yz}{5}=xz\rightarrow x=\frac{y}{5}

From [2] and [3]:

xz=xyy=z\displaystyle xz=xy\rightarrow y=z

sub into the constraint:

5(y/5)+y+y=180\displaystyle 5(y/5)+y+y=180

3y=180\displaystyle 3y=180

y=60\displaystyle y=60

Since y=z, then z=60.

and x=60/5=12

The max is 126060\displaystyle 12\cdot 60\cdot 60

and the constraint is met 5(12)+60+60=180\displaystyle 5(12)+60+60=180
 
Is the following not a theorem?


----> The maximum product of n positive real factors occurs when
all of the factors are equal to each other. <----


*If* this is true, then 5x = y = z,


and then 5x = 60.


So x = 12.


And then the maximum = (12)(60)(60).
 
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