Optimization, Partial Derivatives

**MathTooHard** · October 25th 2009, 01:20 AM

Find three positive numbers x,y,z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum.

At first, I thought a, b, and c needed constant values, but later realized that the numbers are in terms of a, b, and c. Can someone help?

**tonio** · October 25th 2009, 05:07 AM

Originally Posted by MathTooHard

Find three positive numbers x,y,z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum.

At first, I thought a, b, and c needed constant values, but later realized that the numbers are in terms of a, b, and c. Can someone help?

I think this exercise is (almost) impossible to solve without knowing first that relation you say there exists between the variables x,y,z and the powers a,b,c.

Tonio

**Scott H** · October 25th 2009, 05:58 AM

The relations involved are

$x,y>0$
$x+y<100$
$x+y+z=100.$

Our task is to maximize the two-variable function

$u=x^a y^b (100-x-y)^c$

subject to these conditions. In order to do this, we find all critical points, including boundary points and singular points (none, in this case), and points at which $\nabla u=\mathbf{0},$ i.e.,

$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0.$

To begin, we first find the partial derivative of $u$ with respect to $x$ :

$\begin{aligned} \frac{\partial u}{\partial x}&=ax^{a-1}y^b(100-x-y)^c+x^a y^b\cdot c(100-x-y)^{c-1}\cdot(-1)\\ &= ax^{a-1}y^b(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}. \end{aligned}$

Because $x$ and $y$ are symmetric in the function (with $a$ and $b$ respectively), the partial derivative of $u$ with respect to $y$ is therefore

$\frac{\partial u}{\partial y}=bx^a y^{b-1}(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}.$

The critical points are thus all those points at which

$\begin{aligned} ax^{a-1}y^b(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}&=0\\ bx^a y^{b-1}(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}&=0. \end{aligned}$

We may simplify these equations a bit dividing by $x^{a-1}y^{b-1}(100-x-y)^{c-1}$ , remembering that $x$ , $y$ , and $z$ are positive:

$\begin{aligned} ay(100-x-y)-cxy&=0\\ bx(100-x-y)-cxy&=0. \end{aligned}$

Now all that remains is to solve for $x$ and $y$ .

**tonio** · October 25th 2009, 06:38 AM

Originally Posted by Scott H

The relations involved are

$x,y>0$
$x+y<100$
$x+y+z=100.$

Our task is to maximize the two-variable function

$u=x^a y^b (100-x-y)^c$

subject to these conditions. In order to do this, we find all critical points, including boundary points and singular points (none, in this case), and points at which $\nabla u=\mathbf{0},$ i.e.,

$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0.$

To begin, we first find the partial derivative of $u$ with respect to $x$ :

$\begin{aligned} \frac{\partial u}{\partial x}&=ax^{a-1}y^b(100-x-y)^c+x^a y^b\cdot c(100-x-y)^{c-1}\cdot(-1)\\ &= ax^{a-1}y^b(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}. \end{aligned}$

Because $x$ and $y$ are symmetric in the function (with $a$ and $b$ respectively), the partial derivative of $u$ with respect to $y$ is therefore

$\frac{\partial u}{\partial y}=bx^a y^{b-1}(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}.$

The critical points are thus all those points at which

$\begin{aligned} ax^{a-1}y^b(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}&=0\\ bx^a y^{b-1}(100-x-y)^c-cx^a y^b(100-x-y)^{c-1}&=0. \end{aligned}$

We may simplify these equations a bit dividing by $x^{a-1}y^{b-1}(100-x-y)^{c-1}$ , remembering that $x$ , $y$ , and $z$ are positive:

$\begin{aligned} ay(100-x-y)-cxy&=0\\ bx(100-x-y)-cxy&=0. \end{aligned}$

Now all that remains is to solve for $x$ and $y$ .

Wow! What a different looking question with so much more information, and a nice work.
How did you know all this? The OP didn't write it.

Tonio

**Scott H** · October 25th 2009, 06:49 AM

Originally Posted by tonio

Wow! What a different looking question with so much more information, and a nice work.
How did you know all this? The OP didn't write it.

The OP wrote,

"Find three positive numbers x,y,z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum."

This translates into

$x,y>0$
$x+y<100$
$x+y+z=100.$

**tonio** · October 25th 2009, 06:55 AM

Originally Posted by Scott H

The OP wrote,

"Find three positive numbers x,y,z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum."

This translates into

$x,y>0$
$x+y<100$
$x+y+z=100.$

True, but now that I re-read carefully your answer we still don't have any idea what's that relation the OP talks about between a,b,c and the variables x,y,z. This is a must because for derivating we have to take that into account.

In fact, the OP writes "later realized that the numbers are in terms of a, b, and c. Can someone help?", and I assume "the numbers" he talks about are x,y,z.

Tonio

**HallsofIvy** · October 25th 2009, 07:02 AM

Actually, that is exactly what the OP wrote: maximize $x^ay^bz^c$ subject to the conditions x+y+ z= 100, x> 0, y> 0 z> 0 for fixed a, b, c. I confess that I do not understand what he meant by "At first, I thought a, b, and c needed constant values, but later realized that the numbers are in terms of a, b, and c." Yes, the result will depend upon a, b, and c, but that does not mean they are not constants. Perhaps he meant that he thought, at first, that he was to put specific values in for a, b, and c.

Because the conditions x> 0, y> 0, z> 0 make the "feasible region" an open set, I would NOT have done it by looking for etrema on the boundaries. The boundaries are not included in the set (and there may not be a maximum).

Instead of replacing z with 100- x- y, I think I would use the "Lagrange multiplier method". Let $F(x,y,z)= x^ay^bz^c$ . Then $\nabla F= ax^{a-1}y^bz^c\vec{i}+ bx^ay^{b-1}z^c\vec{j}+ cx^ay^bz^{c-1}\vec{k}$ . Let G(x,y,z)= x+ y+ z= 100. Then $\nabla G= \vec{i}+ \vec{j}+\vec{k}$ . At extrema of F, satisfying G= constant, those two vectors must be parallel- one is a multiple of the other: $\nabla F= \lambda\nabla G$ or $ax^{a-1}y^bz^c\vec{i}+ bx^ay^{b-1}z^c\vec{j}+ cx^ay^bz^{c-1}\vec{k}= \lambda(\vec{i}+ \vec{j}+\vec{k})$ .

Equating the components, we have $ax^{a-1}y^bz^c= \lambda$ , $bx^ay^{b-1}z^c= \lambda$ , and $cx^ay^bz^{c-1}= \lambda$ .

Dividing the first equation by the second we get $\frac{a}{b}\frac{y}{x}= 1$ or $y= \frac{b}{a}x$ . Dividing the first equation by the third gives $\frac{a}{c}\frac{z}{x}= 1$ or $z= \frac{c}{a}x$ .

Now put those into x+ y+ z= 100 to get $x+ \frac{b}{a}x+ \frac{c}{a}x= \frac{a+ b+ c}{a} x= 100$ .

$x= \frac{100a}{a+ b+ c}$ . You could put that into $y= \frac{b}{a}x$ and $z= \frac{c}{a}x$ to get that $y= \frac{100b}{a+ b+ c}$ and $z= \frac{100c}{a+ b+ c}$ which follow from the symmetry also.

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