Partial differentials of 2 variables problem

[7777ft]

Problem:
Verify that the sum of three quantities [MATH]x, y, z[/MATH], whose product is a constant [MATH]k[/MATH], is maximum when these three quantities are equal.

My answer [MATH]x=y=z=\sqrt[3]{k}[/MATH] is a minimum according to the second derivative test, and the problem requires a maximum. Are my initial equations incorrect? Are there other critical points?

[MATH]f(x,y,z)=x+y+z[/MATH]
The condition
[MATH]xyz=k \Longrightarrow z=\frac{k}{xy}[/MATH][MATH]f\left(x,y,\frac{k}{xy}\right)=x+y+\frac{k}{xy}=\frac{k + x^2 y + x y^2}{xy}[/MATH]
Partial Derivatives
[MATH]f_{x}=1 - \frac{k}{x^2 y}[/MATH][MATH]f_{xx}=\frac{2 k}{x^3 y}[/MATH][MATH]f_{xy}=\frac{k}{x^2 y^2}[/MATH][MATH]f_{y}=1-\frac{k}{x y^2}[/MATH][MATH]f_{yy}=\frac{2 k}{x y^3}[/MATH]
Set [MATH]f_x=0=1 - \frac{k}{x^2 y}[/MATH]Set [MATH]f_y=0=1 - \frac{k}{x y^2}[/MATH]System of equations resolves to critical point [MATH](x,y)=(\sqrt[3]{k},\sqrt[3]{k})[/MATH] where [MATH]k\neq 0[/MATH][MATH]z=\frac{k}{\sqrt[3]{k}\cdot\sqrt[3]{k}}=\sqrt[3]{k}[/MATH]
Evaluate second order derivatives with critical point
[MATH]f_{xx}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}[/MATH][MATH]f_{yy}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}[/MATH][MATH]f_{xy}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{1}{\sqrt[3]{k}}[/MATH]
Second derivative test
[MATH]D(\sqrt[3]{k},\sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}\cdot\frac{2}{\sqrt[3]{k}}-\left(\frac{1}{\sqrt[3]{k}}\right)^2=\frac{3}{k^{2/3}}[/MATH]........ edited

[MATH][D(\sqrt[3]{k},\sqrt[3]{k}) > 0] \land [f_{xx}(\sqrt[3]{k}, \sqrt[3]{k}) > 0] \therefore[/MATH] critical point is minimum.

 

[Deleted member 4993]

Problem:
Verify that the sum of three quantities [MATH]x, y, z[/MATH], whose product is a constant [MATH]k[/MATH], is maximum when these three quantities are equal.

My answer [MATH]x=y=z=\sqrt[3]{k}[/MATH] is a minimum according to the second derivative test, and the problem requires a maximum. Are my initial equations incorrect? Are there other critical points?

[MATH]f(x,y,z)=x+y+z[/MATH]
The condition
[MATH]xyz=k \Longrightarrow z=\frac{k}{xy}[/MATH][MATH]f\left(x,y,\frac{k}{xy}\right)=x+y+\frac{k}{xy}=\frac{k + x^2 y + x y^2}{xy}[/MATH]
Partial Derivatives
[MATH]f_{x}=1 - \frac{k}{x^2 y}[/MATH][MATH]f_{xx}=\frac{2 k}{x^3 y}[/MATH][MATH]f_{xy}=\frac{k}{x^2 y^2}[/MATH][MATH]f_{y}=1-\frac{k}{x y^2}[/MATH][MATH]f_{yy}=\frac{2 k}{x y^3}[/MATH]
Set [MATH]f_x=0=1 - \frac{k}{x^2 y}[/MATH]Set [MATH]f_y=0=1 - \frac{k}{x y^2}[/MATH]System of equations resolves to critical point [MATH](x,y)=(\sqrt[3]{k},\sqrt[3]{k})[/MATH] where [MATH]k\neq 0[/MATH][MATH]z=\frac{k}{\sqrt[3]{k}\cdot\sqrt[3]{k}}=\sqrt[3]{k}[/MATH]
Evaluate second order derivatives with critical point
[MATH]f_{xx}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}[/MATH][MATH]f_{yy}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}[/MATH][MATH]f_{xy}(\sqrt[3]{k}, \sqrt[3]{k})=\frac{1}{\sqrt[3]{k}}[/MATH]
Second derivative test
[MATH]D(\sqrt[3]{k},\sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}\cdot\frac{2}{\sqrt[3]{k}}-\left(\frac{1}{\sqrt[3]{k}}\right)^2=\frac{2}{k^{2/3}}[/MATH]
[MATH][D(\sqrt[3]{k},\sqrt[3]{k}) > 0] \land [f_{xx}(\sqrt[3]{k}, \sqrt[3]{k}) > 0] \therefore[/MATH] critical point is minimum.

As far as I can see excellent work (I did not review it with 'pencil & paper'). It is all correct - except (I think) there is an inconsequential arithmetic mistake in the second line from the bottom.

I did a quick 'spreadsheet' analysis with k = 27000, and x < 55 and y < 55 and it shows indeed x=y=z (=30 in my example) is a minimum.

Great work and presentation.....

 

[Dr.Peterson]

It would appear that the problem statement is incorrect.

For comparison, consider the case of two variables x and y. Their product is constant along a rectangular hyperbola xy=k, and clearly their sum is a minimum, not a maximum, when they are equal (the vertex of the hyperbola). For example, if xy=1, then when x=y=1, the sum is 1+1=2; when x=2 and y=1/2, the sum is x+y = 2 1/2, which is greater.

The same is true of your problem, as SK has demonstrated, and as you can show by similar examples. Your conclusion is correct.

 

[7777ft]

there is an inconsequential arithmetic mistake in the second line from the bottom.

You're right. I can't find a button to correct my post. At any rate the second line from the bottom of post #1 should be
[MATH]D(\sqrt[3]{k},\sqrt[3]{k})=\frac{2}{\sqrt[3]{k}}\cdot\frac{2}{\sqrt[3]{k}}-\left(\frac{1}{\sqrt[3]{k}}\right)^2=\frac{3}{k^{2/3}}[/MATH]
I can't seem to change the title either; something along the lines of "Maximize the sum of 3 equal values" would've been more descriptive.