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.
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.
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