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(D) The constraints are $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x = y$,and $x, y \geqslant 0$. Substituting $x = y$ into the constraints: $1$) $x

Vedclass · Accepted Answer

$80$

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(D) The constraints are $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x = y$,and $x, y \geqslant 0$. Substituting $x = y$ into the constraints: $1$) $x + 3x \leqslant 60 \implies 4x \leqslant 60 \implies x \leqslant 15$. $2$) $x + x \geqslant 10 \implies 2x \geqslant 10 \implies x \geqslant 5$. Since $x = y$,the feasible region is the line segment from $(5, 5)$ to $(15, 15)$. The objective function is $z = 3x + 5y$. Substituting $y = x$ into $z$,we get $z = 3x + 5x = 8x$. At $(5, 5)$,$z = 8(5) = 40$. At $(15, 15)$,$z = 8(15) = 120$. The maximum value is $120$ and the minimum value is $40$. The difference is $120 - 40 = 80$.

The difference between the maximum value and minimum value of the objective function $z = 3x + 5y$ subject to the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y = 0$,and $x, y \geqslant 0$ is:

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