A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank

Q: Wait, why don't we need to know the length of the river bank to solve this?

That is a brilliant observation! In many geometry problems, we are conditioned to find every side length before calculating the area. However, in this optimization problem, the river bank is a 'dependent' variable. As you change the angle $\theta$ between the two fences, the distance between the two endpoints (the river bank) changes automatically to close the triangle. We don't need to calculate that length because the area formula $A = \frac{1}{2}ab \sin(\theta)$ uses the two sides we already know. The river bank is just the boundary that completes the shape; it doesn't constrain the area calculation itself.

Q: Can I use calculus to solve this instead of just knowing the maximum of the sine function?

Absolutely! Using calculus is a fantastic way to verify your intuition. If you define $A(\theta) = \frac{1}{2}x^2 \sin(\theta)$, you can find the derivative with respect to $\theta$: $\frac{dA}{d\theta} = \frac{1}{2}x^2 \cos(\theta)$. Setting this derivative to zero for a maximum gives $\cos(\theta) = 0$, which implies $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians. This confirms our trigonometric intuition perfectly. Calculus is the 'heavy artillery' of physics and math—it's always there if you need to be rigorous!

Q: Why is the maximum area $\frac{1}{2}x^2$? Does this make sense physically?

Think of it this way: when $\theta = 90^\circ$, the two fences form the base and the height of a right-angled triangle. The area of a right triangle is simply $\frac{1}{2} \times \text{base} \times \text{height}$. Since both the base and height are $x$, the area is $\frac{1}{2} \times x \times x = \frac{1}{2}x^2$. It is the most 'efficient' way to arrange those two fences to capture space. Any angle smaller than $90^\circ$ makes the triangle 'skinny,' and any angle larger than $90^\circ$ makes it 'squashed,' both reducing the area.

Question

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $x$. The maximum area enclosed by the park is

Options:

(A) $\frac{3}{2} x^2$

(B) $\sqrt{\frac{x^3}{8}}$

(C) $\frac{1}{2} x^2$

(D) $\pi x^2$

Sigma Percentile · Accepted Answer

### The Geometry of Efficiency: Maximizing Your Park

Imagine you are standing on the bank of a serene, straight river. You have been tasked with creating a triangular park. You have two fences, each of length $x$, and you need to arrange them to enclose the largest possible area.

The river bank will serve as the third side of your triangle. This is not just a math problem; it is a fundamental question about efficiency and the nature of space. Let us walk through this journey together.

### Phase 1: The Canvas of the Problem

When we approach a problem like this, the first step is always to visualize the constraints. We have a fixed length $x$ for two sides. Let us denote the angle between these two fences as $	heta$.

As you stand at the vertex where the two fences meet, you can imagine swinging the fences open or closed. If you bring them close together, the area is tiny. If you open them wide, the area grows.

But there is a limit. The river bank is a straight line, and as you change $	heta$, the distance between the ends of your fences changes to accommodate the shape. We are looking for the 'sweet spot'—the angle that captures the most land.

### Phase 2: The Mathematical Tool

To solve this, we need a bridge between geometry and algebra. We need a formula that relates the area of a triangle to the sides we know.

You might remember the standard formula $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. But here, we do not know the height directly. Instead, we have the two sides $x$ and $x$, and the included angle $	heta$.

This is where the trigonometric area formula becomes our best friend:

$$A = \frac{1}{2} ab \sin(	heta)$$

In our specific scenario, both $a$ and $b$ are equal to $x$. When we substitute these into our equation, we get:

$$A = \frac{1}{2} (x)(x) \sin(	heta) = \frac{1}{2} x^2 \sin(	heta)$$

Look at this equation. It is elegant, is it not? The term $\frac{1}{2} x^2$ is a constant—it is fixed by the length of your fences. The only thing that can change, the only 'knob' we can turn to influence the area, is $\sin(	heta)$.

### Phase 3: The Optimization

Now, we enter the realm of optimization. We want to maximize $A$. Since $\frac{1}{2} x^2$ is positive and constant, the area $A$ is maximized precisely when $\sin(	heta)$ is at its maximum.

Ask yourself: what is the maximum value of the sine function? If you recall your unit circle, the sine function oscillates between $-1$ and $1$. Therefore, the absolute maximum value it can ever reach is $1$.

So, we set our goal: we need $\sin(	heta) = 1$.

When does this happen? It happens when $	heta = 90^\circ$ (or $\frac{\pi}{2}$ radians). This is a profound geometric realization. It tells us that to enclose the maximum area with two fences of equal length, they must be perpendicular to each other.

Nature loves symmetry and right angles; they are often the markers of optimal efficiency.

### Phase 4: The Final Calculation

With the optimal angle identified, the rest is simple arithmetic. We substitute $\sin(	heta) = 1$ back into our area equation:

$$A_{max} = \frac{1}{2} x^2 (1) = \frac{1}{2} x^2$$

And there it is. The **maximum area is $\frac{1}{2} x^2$**.

### Reflection

Think about what we just did. We took a physical scenario—a park by a river—and translated it into the language of trigonometry. We identified the variable, we found the function, and we used the properties of the sine function to find the peak of that function.

This is the essence of JEE Advanced physics and mathematics. It is not about memorizing formulas; it is about understanding how variables interact.

Whenever you face an optimization problem in the future, remember this park. Ask yourself: 'What is the variable I can control?' and 'What is the function that describes the outcome?' Once you have those two, the path to the solution will always reveal itself. You have done excellent work today. Keep that curiosity alive!

Animated Solution for Mathematics - Differentiation: A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $x$ . The maximum area enclosed by the park is

Select Answer:

Visualized Solution

Visualizing the Park

The Area Formula

Substituting Known Values

Simplifying the Expression

Maximizing the Area

Finding the Optimal Angle

Calculating Maximum Area

Final Conclusion

The Sigma Insight: Maxima and Minima

The Geometry of Efficiency

Phase 1

Phase 2

Phase 3

Phase 4

Reflection

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