Key Concepts and Formulas

• Linear Programming: Optimization of a linear objective function subject to linear constraints.
• Feasible Region: The region defined by the set of linear constraints.
• Extreme Value Theorem: A continuous function on a closed and bounded set attains its maximum and minimum values.
• Quadratic Function Vertex: For a quadratic function $f(x) = ax^2 + bx + c$, the vertex occurs at $x = -\frac{b}{2a}$.

Step 1: Define the Feasible Region

The constraints are:

1. $3x + 4y \le 100$
2. $4x + 3y \le 75$
3. $x \ge 0$
4. $y \ge 0$

The conditions $x \ge 0$ and $y \ge 0$ restrict the feasible region to the first quadrant. We need to find the vertices of the region defined by these inequalities.

• Line L1: $3x + 4y = 100$

  • $x$-intercept: When $y=0$, $3x = 100 \implies x = \frac{100}{3}$. Point: $(\frac{100}{3}, 0)$.
  • $y$-intercept: When $x=0$, $4y = 100 \implies y = 25$. Point: $(0, 25)$.

• Line L2: $4x + 3y = 75$

  • $x$-intercept: When $y=0$, $4x = 75 \implies x = \frac{75}{4}$. Point: $(\frac{75}{4}, 0)$.
  • $y$-intercept: When $x=0$, $3y = 75 \implies y = 25$. Point: $(0, 25)$.

Now, we find the intersection point of the two lines $3x + 4y = 100$ and $4x + 3y = 75$. Multiply the first equation by 4 and the second by 3:

$12x + 16y = 400 \quad (1)$ $12x + 9y = 225 \quad (2)$

Subtract equation (2) from equation (1): $7y = 175 \implies y = 25$

Substitute $y = 25$ into $3x + 4y = 100$: $3x + 4(25) = 100 \implies 3x + 100 = 100 \implies 3x = 0 \implies x = 0$

So, the intersection point is $(0, 25)$.

The vertices of the feasible region are:

• $O = (0, 0)$
• $A = (\frac{75}{4}, 0)$
• $B = (0, 25)$
• $C = (x,y)$, where $x$ and $y$ satisfy both equations simultaneously.

Since the intersection point is (0,25), and both lines intersect the y-axis at (0,25), we have to solve for the intersection of the two lines.

To find the intersection point of $3x + 4y = 100$ and $4x + 3y = 75$, we multiply the first equation by 3 and the second by 4: $9x + 12y = 300$ $16x + 12y = 300$ Subtracting the first equation from the second gives $7x = 0$, so $x=0$. Substituting $x=0$ into the first equation gives $4y=100$, so $y=25$. The intersection point is $(0, 25)$.

The feasible region is a quadrilateral with vertices at (0,0), (75/4, 0), (0, 25) and the intersection of the two lines. Solving for the intersection of $3x+4y=100$ and $4x+3y=75$ gives us: Multiply the first equation by 4 and the second by 3: $12x+16y=400$ $12x+9y=225$ Subtract the second equation from the first to get $7y=175$, so $y=25$. Plugging this into the first equation, $3x+4(25)=100$, so $3x=0$ and $x=0$. So the intersection point is (0,25).

We need to find where the two lines intersect. Multiply $3x+4y=100$ by 4 to get $12x+16y=400$. Multiply $4x+3y=75$ by 3 to get $12x+9y=225$. Subtracting gives $7y=175$, so $y=25$. Then $3x+4(25)=100$, so $3x=0$ and $x=0$. Thus the lines intersect at $(0,25)$.

The vertices of the feasible region are O(0,0), A(75/4, 0), and B(0, 25). Let's find the intersection of $3x+4y=100$ and $4x+3y=75$. Multiply the first equation by 4 and the second by -3 to get: $12x+16y=400$ $-12x-9y=-225$ Adding these gives $7y=175$, so $y=25$. Then $3x+4(25)=100$, so $3x=0$ and $x=0$. So the intersection point is $(0,25)$.

Multiplying $3x+4y=100$ by 3 and $4x+3y=75$ by 4, we get $9x+12y=300$ $16x+12y=300$ Subtracting the first from the second, we get $7x=0$, so $x=0$, so $y=25$.

Now multiply the first equation by 4 and the second by 3. $12x+16y=400$ $12x+9y=225$ Subtracting yields $7y = 175$ or $y=25$. Plugging this into the first equation yields $3x+100=100$ or $x=0$. So the intersection point is (0,25).

The vertices are (0,0), (75/4, 0) and (0, 25).

Step 2: Evaluate the Objective Function at the Vertices

The objective function is $z = 6xy + y^2$.

• At $O(0, 0)$: $z = 6(0)(0) + (0)^2 = 0$

• At $A(\frac{75}{4}, 0)$: $z = 6(\frac{75}{4})(0) + (0)^2 = 0$

• At $B(0, 25)$: $z = 6(0)(25) + (25)^2 = 0 + 625 = 625$

Step 3: Evaluate the Objective Function Along the Edges

• Edge OA ($y = 0$, for $0 \le x \le \frac{75}{4}$): Substitute $y = 0$ into $z$: $z = 6x(0) + (0)^2 = 0$

• Edge OB ($x = 0$, for $0 \le y \le 25$): Substitute $x = 0$ into $z$: $z = 6(0)y + y^2 = y^2$ The maximum value on this edge is $25^2 = 625$ at $y = 25$.

• Edge AB ($4x + 3y = 75$): Express $x$ in terms of $y$: $4x = 75 - 3y \implies x = \frac{75 - 3y}{4}$ Substitute into $z$: $z = 6\left(\frac{75 - 3y}{4}\right)y + y^2 = \frac{3}{2}(75 - 3y)y + y^2 = \frac{225y - 9y^2}{2} + y^2 = \frac{225y - 9y^2 + 2y^2}{2} = \frac{225y - 7y^2}{2}$ Let $f(y) = \frac{225y - 7y^2}{2}$. To find the maximum, take the derivative and set it to zero: $f'(y) = \frac{225 - 14y}{2} = 0 \implies 225 - 14y = 0 \implies y = \frac{225}{14}$ Since $0 \le y \le 25$, and $\frac{225}{14} \approx 16.07 < 25$, this is a valid critical point. Then $x = \frac{75 - 3(\frac{225}{14})}{4} = \frac{75 - \frac{675}{14}}{4} = \frac{\frac{1050 - 675}{14}}{4} = \frac{375}{56}$ Now, substitute $y = \frac{225}{14}$ into $z$: $z = \frac{225(\frac{225}{14}) - 7(\frac{225}{14})^2}{2} = \frac{\frac{225^2}{14} - \frac{7 \cdot 225^2}{14^2}}{2} = \frac{\frac{225^2}{14} - \frac{225^2}{28}}{2} = \frac{\frac{225^2}{28}}{2} = \frac{225^2}{56} = \frac{50625}{56} \approx 904.02$

Step 4: Determine the Overall Maximum Value

Comparing the values of $z$ at the vertices and along the edges, we have:

• $z(0, 0) = 0$
• $z(\frac{75}{4}, 0) = 0$
• $z(0, 25) = 625$
• $z(\frac{375}{56}, \frac{225}{14}) = \frac{50625}{56} \approx 904.02$

The maximum value is $\frac{50625}{56}$.

Common Mistakes & Tips

• Always check the function's behavior along the edges of the feasible region, as the maximum may not occur at a vertex.
• Be careful when calculating the intersection points of the constraint lines.
• Double-check your algebra when substituting and simplifying expressions.

Summary

We found the feasible region defined by the given constraints and evaluated the objective function at the vertices. Then, we analyzed the objective function along the edges by expressing one variable in terms of the other and finding critical points. Comparing the values obtained, we found that the maximum value of the objective function is $\frac{50625}{56}$.

The final answer is $\boxed{\frac{50625}{56}}$.