Animated Solution for Mathematics - Differentiation: The maximum value of $z$ in the following equation $z=6xy+y^2$, where $3x+4y\le 100$ and $4x+3y\le 75$ for $x\ge 0$ and $y\ge 0$ is ____ (Round off to the Nearest Integer)

Finding the Peak

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving an equation; we are embarking on a journey of optimization.

We are given an objective function, $z=6xy+y^2$, and we want to find its absolute maximum value. We are bound by the constraints $3x+4y\le 100$ and $4x+3y\le 75$, with the added condition that $x,y\ge 0$.

This is the classic setup of a constrained optimization problem, and it is a beautiful dance between algebra and geometry.

The Landscape of Constraints

Imagine you are standing in the first quadrant of the Cartesian plane. You have two fences, $3x+4y=100$ and $4x+3y=75$. These fences define your 'feasible region'—the only place where you are allowed to exist.

If you plot these lines, you will notice something fascinating. Both lines intersect the $y$-axis at $(0,25)$. However, the second line, $4x+3y=75$, has a smaller $x$-intercept than the first.

This means the second line is 'tighter' and encloses a smaller area. Any point that satisfies $4x+3y\le 75$ will automatically satisfy $3x+4y\le 100$. Thus, the first constraint is redundant, and we can focus entirely on the boundary $4x+3y=75$.

The Reduction

We have a function $z=6xy+y^2$ and a constraint $4x+3y=75$. We want to maximize $z$. Since $z$ increases as $x$ and $y$ increase, we know the maximum must lie on the boundary.

Let's reduce this two-variable problem to a single-variable one. From our active constraint, we can isolate $x$:

This is our key. By substituting this into our objective function, we transform the problem into a simple quadratic function of $y$.

The Calculus of Peaks

Let's perform the substitution: $z=y(6x+y)$. Substituting $x$, we get:

Simplifying this, we obtain:

Now, we have a function $f(y)=\frac{225y-7y^2}{2}$. This is a downward-opening parabola. To find its peak, we take the derivative with respect to $y$ and set it to zero:

Solving this gives $y=\frac{225}{14}\approx 16.07$. This is the $y$-coordinate of our maximum.

The Grand Finale

With $y=\frac{225}{14}$, we find the corresponding $x$ using $x=\frac{75-3y}{4}$, which yields:

We verify this point against our constraints, and it fits perfectly. Finally, we calculate the maximum value:

Rounding to the nearest integer, we arrive at 904. You have successfully navigated the constraints and found the peak! Remember, in JEE, it is not just about the calculation; it is about understanding the geometry behind the algebra.