The corner points of the bounded feasible region determined by the system of linear constraints are $(0, 0), (5, 0), (6, 5), (6, 8), (4, 10), (0,8)$. Let $z = 3x - 4y$ be the objective function. Then the minimum value of the objective function z occurs at

In linear programming, the minimum or maximum value of the objective function always occurs at one of the corner points of the feasible region.

What does this mean? Instead of checking every possible point in the region, we only need to test the corner points - this saves us tons of time!

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To find the minimum value of $z = 3x - 4y$, calculate $z$ at each corner point:

At $(0, 0)$:

$z = 3(0) - 4(0) = 0$

At $(5, 0)$:

$z = 3(5) - 4(0) = 15$

At $(6, 5)$:

$z = 3(6) - 4(5) = 18 - 20 = -2$

At $(6, 8)$:

$z = 3(6) - 4(8) = 18 - 32 = -14$

At $(4, 10)$:

$z = 3(4) - 4(10) = 12 - 40 = -28$

At $(0, 8)$:

$z = 3(0) - 4(8) = 0 - 32 = -32$

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Comparing all values: $-32 < -28 < -14 < -2 < 0 < 15$

Remember: For minimum, we want the smallest (most negative) number!

The smallest value is $z = -32$.

Therefore, the minimum value of the objective function occurs at $(0, 8)$.

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Quick Tip (Pattern Recognition):

The objective function $z = 3x - 4y$ has $-4$ multiplying $y$. Since this is negative and larger in magnitude than the coefficient of $x$ (which is $3$), larger values of $y$ will make $z$ more negative (smaller).

So look for corner points with the biggest $y$-values first - they're likely to give you the minimum!