Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If AD = 8m, BC =

Q: Why do we use the coordinate geometry approach instead of just using simple geometry?

While simple geometry can sometimes work, coordinate geometry is a powerful, systematic tool for optimization problems. By assigning coordinates, we transform a geometric visualization into a clear algebraic function, $S(x)$, which allows us to use the robust machinery of calculus to find the exact minimum.

Q: Why are we minimizing the sum of squares, $MD^2 + MC^2$, instead of just the sum of distances, $MD + MC$?

The problem specifically asks to minimize $MD^2 + MC^2$. Minimizing the sum of squares is mathematically much cleaner because it results in a quadratic function, which is easy to differentiate. Minimizing the sum of distances, $MD + MC$, would involve square roots, leading to a much more complex optimization problem.

Q: How do we know that the value we found is a minimum and not a maximum?

Great question! We use the second derivative test. After finding the first derivative $S'(x) = 4x - 20$ and setting it to zero to get $x=5$, we calculate the second derivative $S''(x)$. Since $S''(x) = 4$, which is positive, the function is concave up, confirming that $x=5$ is indeed a local minimum.

Question

Let $AD$ and $BC$ be two vertical poles at $A$ and $B$ respectively on a horizontal ground. If $AD = 8	ext{m}$, $BC = 11	ext{m}$ and $AB = 10	ext{m}$; then the distance (in meters) of a point $M$ on $AB$ from the point $A$ such that $MD^2 + MC^2$ is minimum is

Sigma Percentile · Accepted Answer

### The Geometry of Optimization

Imagine you are standing on a flat, horizontal ground. You see two vertical poles, $AD$ and $BC$, rising like sentinels.

Pole $AD$ stands $8$ meters tall, and pole $BC$ stands $11$ meters tall. They are separated by a distance of $10$ meters along the ground.

You are tasked with finding a point $M$ on the ground between these two poles such that the sum of the squares of the distances from $M$ to the tops of the poles, $D$ and $C$, is minimized. This is not just a math problem; it is a challenge of finding the perfect balance in a physical system.

### Phase 1: Setting the Stage

To solve this, we must first translate the physical reality into the language of mathematics. Let us define our coordinate system.

We place the base of the first pole, point $A$, at the origin $(0,0)$. Since the second pole is $10$ meters away, point $B$ sits at $(10,0)$.

Now, we can easily define the tops of the poles. Point $D$ is at $(0,8)$, and point $C$ is at $(10,11)$. By setting up this coordinate system, we have turned a physical scene into a precise geometric map.

### Phase 2: The Algebraic Dance

Now, let us place point $M$ on the line segment $AB$. Let its distance from $A$ be $x$ meters.

Thus, the coordinates of $M$ are $(x,0)$, where $0 \le x \le 10$. We want to minimize the sum of the squares of the distances $MD$ and $MC$.

Using the distance formula, we find:

$$MD^2 = (x - 0)^2 + (0 - 8)^2 = x^2 + 64$$

Similarly, for the second pole:

$$MC^2 = (x - 10)^2 + (0 - 11)^2 = (x - 10)^2 + 121$$

We define our function $S(x)$ as the sum of these squares:

$$S(x) = MD^2 + MC^2 = (x^2 + 64) + ((x - 10)^2 + 121)$$

Expanding the term $(x - 10)^2$ gives us $x^2 - 20x + 100$. Substituting this back into our function, we get:

$$S(x) = x^2 + 64 + x^2 - 20x + 100 + 121$$

Combining like terms, we arrive at a beautiful, simplified quadratic function:

$$S(x) = 2x^2 - 20x + 285$$

### Phase 3: The Calculus Moment

Now, we enter the realm of calculus. To find the minimum of this function, we look for the point where the slope of the curve is zero.

We take the first derivative of $S(x)$ with respect to $x$:

$$S'(x) = \frac{d}{dx}(2x^2 - 20x + 285) = 4x - 20$$

Setting $S'(x) = 0$ to find the critical point, we have:

$$4x - 20 = 0 \Rightarrow 4x = 20 \Rightarrow x = 5$$

This tells us that the potential minimum occurs exactly $5$ meters from point $A$. To be absolutely certain this is a minimum, we check the second derivative:

$$S''(x) = \frac{d}{dx}(4x - 20) = 4$$

Since $S''(x) = 4 > 0$, the function is concave up, which mathematically guarantees that **$x=5$** is a local minimum.

The point $M$ must be placed exactly **$5$ meters from $A$** to minimize the sum of the squared distances. It is a simple, elegant result for a problem that initially seemed complex.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Defining Coordinates of Poles

Introducing Point $M$

The Distance Formula for $M D^{2}$

The Distance Formula for $M C^{2}$

Defining the Function $S (x)$

Expanding the Expression

Simplifying the Function

The Condition for Minima

Differentiating the Function

Solving for $x$

The Second Derivative Test

Final Result

The Sigma Insight: Maxima and Minima

The Geometry of Optimization

Phase 1

Phase 2

Phase 3

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