The maximum value of z in the following equation z = 6xy + y2, where 3x + 4y \le 100 and 4x + 3y \le 75 for x \ge 0 and y \ge 0 is \underline\hspace2cm .

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q39 with Solution

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Correct answer: 50625/56

We need to maximize $z=6xy+y^2$ subject to $3x+4y\le 100,\qquad 4x+3y\le 75,\qquad x\ge 0,\ y\ge 0.$
Since the feasible region is determined by linear inequalities, let us first find its corner points.

On the $x$ -axis ( $y=0$ ): $3x\le 100,\quad 4x\le 75 \implies x\le \frac{75}{4}.$ So one vertex is $\left(\frac{75}{4},0\right).$
On the $y$ -axis ( $x=0$ ): $4y\le 100,\quad 3y\le 75 \implies y\le 25.$ So another vertex is $(0,25).$
The origin $(0,0)$ is also feasible.
Intersection of $3x+4y=100,\qquad 4x+3y=75.$ Solve:

Multiply the first by $4$ and the second by $3$ : $12x+16y=400,$ $12x+9y=225.$ Subtract: $7y=175 \implies y=25.$ Then $4x+3(25)=75 \implies 4x=0 \implies x=0.$ So these two lines intersect at $(0,25)$ .

Hence the feasible region is the triangle with vertices $(0,0),\quad \left(\frac{75}{4},0\right),\quad (0,25).$

Observe that $z=6xy+y^2=y(6x+y).$ For fixed $y$ , $z$ increases as $x$ increases because the coefficient of $x$ is $6y\ge 0$ . Thus for a given $y$ , we should take the largest possible $x$ .

From the constraints, $x\le \frac{100-4y}{3},\qquad x\le \frac{75-3y}{4}.$ Now compare these two bounds:

=\frac{400-16y-225+9y}{12} =\frac{175-7y}{12}.

Since in the feasible region $0\le y\le 25$ , we have $175-7y\ge 0$ . Therefore, $\frac{75-3y}{4}\le \frac{100-4y}{3}.$ So the active constraint is $x=\frac{75-3y}{4}.$

Substitute this into $z$ :

=\frac{6y(75-3y)}{4}+y^2 =\frac{450y-18y^2}{4}+y^2.

Simplify:

=\frac{225y-7y^2}{2}.$$ So we maximize $$f(y)=\frac{225y-7y^2}{2}, \qquad 0\le y\le 25.$$ 5. This is a downward-opening parabola. Its vertex occurs at $$y=\frac{-b}{2a}$$ for $ay^2+by+c$. Here, $$f(y)=\frac{1}{2}(-7y^2+225y),$$ so the maximizing value is $$y=\frac{225}{14}.$$ This lies in $[0,25]$, so it gives the maximum. Then $$x=\frac{75-3y}{4}= rac{75-3\cdot \frac{225}{14}}{4} =\frac{\frac{1050-675}{14}}{4} =\frac{375}{56}.$$ 6. Now compute the maximum value: $$z_{\max}=6xy+y^2 =6\left(\frac{375}{56}\right)\left(\frac{225}{14}\right)+\left(\frac{225}{14}\right)^2.$$ It is easier to use $$z=\frac{225y-7y^2}{2}.$$ Substitute $y=\frac{225}{14}$: $$z_{\max}=\frac{1}{2}\left(225\cdot \frac{225}{14}-7\cdot \frac{225^2}{14^2}\right).$$ $$=\frac{1}{2}\left(\frac{225^2}{14}-\frac{225^2}{28}\right) =\frac{1}{2}\left(\frac{225^2}{28}\right) =\frac{225^2}{56}.

Now $225^2=50625,$ so $z_{\max}=\frac{50625}{56}\approx 904.017857.$

Therefore, the maximum value is $\boxed{\frac{50625}{56}}.$ Since the question is of integer type, this does not equal the stored integer answer $904$ exactly. If the exam intended the greatest integer value, then it would be $904$ , but the mathematical maximum is $\frac{50625}{56}.$

More from Straight Lines and Pair of Straight Lines

View written solutionFree

Correct answer: 50625/56

We need to maximize $z=6xy+y^2$ subject to $3x+4y\le 100,\qquad 4x+3y\le 75,\qquad x\ge 0,\ y\ge 0.$
Since the feasible region is determined by linear inequalities, let us first find its corner points.

On the $x$ -axis ( $y=0$ ): $3x\le 100,\quad 4x\le 75 \implies x\le \frac{75}{4}.$ So one vertex is $\left(\frac{75}{4},0\right).$
On the $y$ -axis ( $x=0$ ): $4y\le 100,\quad 3y\le 75 \implies y\le 25.$ So another vertex is $(0,25).$
The origin $(0,0)$ is also feasible.
Intersection of $3x+4y=100,\qquad 4x+3y=75.$ Solve:

Multiply the first by $4$ and the second by $3$ : $12x+16y=400,$ $12x+9y=225.$ Subtract: $7y=175 \implies y=25.$ Then $4x+3(25)=75 \implies 4x=0 \implies x=0.$ So these two lines intersect at $(0,25)$ .

Hence the feasible region is the triangle with vertices $(0,0),\quad \left(\frac{75}{4},0\right),\quad (0,25).$

Observe that $z=6xy+y^2=y(6x+y).$ For fixed $y$ , $z$ increases as $x$ increases because the coefficient of $x$ is $6y\ge 0$ . Thus for a given $y$ , we should take the largest possible $x$ .

From the constraints, $x\le \frac{100-4y}{3},\qquad x\le \frac{75-3y}{4}.$ Now compare these two bounds:

=\frac{400-16y-225+9y}{12} =\frac{175-7y}{12}.

Since in the feasible region $0\le y\le 25$ , we have $175-7y\ge 0$ . Therefore, $\frac{75-3y}{4}\le \frac{100-4y}{3}.$ So the active constraint is $x=\frac{75-3y}{4}.$

Substitute this into $z$ :

=\frac{6y(75-3y)}{4}+y^2 =\frac{450y-18y^2}{4}+y^2.

Simplify:

=\frac{225y-7y^2}{2}.$$ So we maximize $$f(y)=\frac{225y-7y^2}{2}, \qquad 0\le y\le 25.$$ 5. This is a downward-opening parabola. Its vertex occurs at $$y=\frac{-b}{2a}$$ for $ay^2+by+c$. Here, $$f(y)=\frac{1}{2}(-7y^2+225y),$$ so the maximizing value is $$y=\frac{225}{14}.$$ This lies in $[0,25]$, so it gives the maximum. Then $$x=\frac{75-3y}{4}= rac{75-3\cdot \frac{225}{14}}{4} =\frac{\frac{1050-675}{14}}{4} =\frac{375}{56}.$$ 6. Now compute the maximum value: $$z_{\max}=6xy+y^2 =6\left(\frac{375}{56}\right)\left(\frac{225}{14}\right)+\left(\frac{225}{14}\right)^2.$$ It is easier to use $$z=\frac{225y-7y^2}{2}.$$ Substitute $y=\frac{225}{14}$: $$z_{\max}=\frac{1}{2}\left(225\cdot \frac{225}{14}-7\cdot \frac{225^2}{14^2}\right).$$ $$=\frac{1}{2}\left(\frac{225^2}{14}-\frac{225^2}{28}\right) =\frac{1}{2}\left(\frac{225^2}{28}\right) =\frac{225^2}{56}.

Now $225^2=50625,$ so $z_{\max}=\frac{50625}{56}\approx 904.017857.$

Therefore, the maximum value is $\boxed{\frac{50625}{56}}.$ Since the question is of integer type, this does not equal the stored integer answer $904$ exactly. If the exam intended the greatest integer value, then it would be $904$ , but the mathematical maximum is $\frac{50625}{56}.$

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