How do you maximize 3x+4y-yz, subject to x+y<4, z>3?

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How do you maximize 3x+4y-yz, subject to x+y<4, z>3?

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Answer 1 · 2016-05-22T18:47:15

$\infty$ $infty$

Explanation:

$f (x, y, z) = 3 x + 4 y - y z$ $f(x,y,z)=3x+4y-y z$ has not stationary points because there are no points obeying the condition

$\nabla f (x, y, z) = \to 0$ $grad f(x,y,z) = vec 0$

So their extrema could be located at the viable region frontiers. Taking a restriction frontier, for instance $g_{2} (x, y, z) = z - 3 = 0$ $g_2(x,y,z)=z-3=0$ and substituting in $f (x, y, z)$ $f(x,y,z)$ we get

${f (x, y, z)}_{g_{2}} = f_{g_{2}} (x, y) = 3 x + y$ $f(x,y,z)_{g_2} = f_{g_2}(x,y)=3 x + y$

calculating $\nabla f_{g_{2}} (x, y) = {3, 1}$ $grad f_{g_2}(x,y) = {3,1}$
so also no stationary points over $z = 3$ $z=3$

The reduced problem reads now
Maximize $f_{g_{2}} (x, y) = 3 x + y$ $f_{g_2}(x,y)=3 x + y$ with the border restriction
$x + y = 4$ $x+y=4$ . Applying the same idea as before, substituting the border relation in the objective function, we attain
${(f_{g_{2}})}_{g_{1}} = 3 x + (4 - x) = 2 x + 4$ $(f_{g_2})_{g_1} = 3x+(4-x)=2x+4$ we see that the value range for ${(f_{g_{2}})}_{g_{1}}$ $(f_{g_2})_{g_1}$ is unlimited

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How do you maximize 3x+4y-yz, subject to x+y<4, z>3?

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