Two vertical light poles of height \(22 \text{ m}\) and \(16 \text{ m}\) stand on the opposite sides of a \(20 \text{ m}\) wide road.
Two ladders of length \(l_1\) and \(l_2\) are placed from a common point \(R\) on the road at a distance of \(x \text{ m}\) from the smaller pole.
Based on the above information,
Question: 1
Express \(p(x) = l_1 + l_2\) in terms of \(x\).
Show Hint
When a point is placed between two objects, always express the two segments as \(x\) and \((\text{Total} - x)\).
This ensures the function is expressed in terms of a single variable, making it ready for differentiation.
Concept: • Use the Pythagorean theorem for right-angled triangles to find the length of the hypotenuse.
• If a point \(R\) divides a road of width \(W\) into two segments \(x\) and \(W-x\), the distance from \(R\) to the top of a pole of height \(h\) is \(\sqrt{h^2 + \text{segment}^2}\).
Step 1: Identify the dimensions for ladder \(l_2\)
Ladder \(l_2\) connects point \(R\) to the top of the smaller pole (height \(16 \text{ m}\)).
The horizontal distance is given as \(x\).
Using Pythagoras theorem:
\[ l_2 = \sqrt{16^2 + x^2} = \sqrt{256 + x^2} \]
Step 2: Identify the dimensions for ladder \(l_1\)
Ladder \(l_1\) connects point \(R\) to the top of the larger pole (height \(22 \text{ m}\)).
The total road width is \(20 \text{ m}\).
Since \(R\) is \(x \text{ m}\) from the smaller pole, its distance from the larger pole is \((20 - x) \text{ m}\).
Using Pythagoras theorem:
\[ l_1 = \sqrt{22^2 + (20 - x)^2} = \sqrt{484 + (20 - x)^2} \]
Step 3: Combine the expressions to find \(p(x)\)
The total length function is the sum of the two ladder lengths:
\[ p(x) = \sqrt{484 + (20 - x)^2} + \sqrt{256 + x^2} \]
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Question: 2
Find \(p'(x)\).
Show Hint
Be careful with the internal derivative of \((20-x)^2\); the negative sign from the coefficient of \(x\) must be accounted for.
Simplifying the fractions by cancelling the factor of 2 makes the expression easier to work with.
• Use the chain rule for differentiation: \(\frac{d}{dx}[\sqrt{u}] = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}\).
• Apply the power rule and sum rule where necessary.
Step 1: Differentiate the first term
Let \(u = 484 + (20 - x)^2\).
Then \(\frac{du}{dx} = 0 + 2(20 - x)(-1) = -2(20 - x)\).
Derivative of the first term:
\[ \frac{d}{dx} \sqrt{484 + (20 - x)^2} = \frac{-2(20 - x)}{2\sqrt{484 + (20 - x)^2}} = \frac{-(20 - x)}{\sqrt{484 + (20 - x)^2}} \] Step 2: Differentiate the second term
Let \(v = 256 + x^2\).
Then \(\frac{dv}{dx} = 2x\).
Derivative of the second term:
\[ \frac{d}{dx} \sqrt{256 + x^2} = \frac{2x}{2\sqrt{256 + x^2}} = \frac{x}{\sqrt{256 + x^2}} \] Step 3: Combine the results
\[ p'(x) = \frac{x}{\sqrt{256 + x^2}} - \frac{20 - x}{\sqrt{484 + (20 - x)^2}} \]
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Question: 3
Find the value of \(x\) for which \(l_1^2 + l_2^2\), is minimum.
Show Hint
Minimizing the sum of squares is often much simpler than minimizing the sum itself because it eliminates square root symbols.
Always check the second derivative to ensure the critical point is a minimum.
Concept: • To minimize a function \(S(x)\), find its derivative \(S'(x)\), set it to zero, and solve for \(x\).
• Use the second derivative test \(S''(x) > 0\) to confirm a local minimum.
Step 1: Define the function \(S(x)\)
From part (i), we have \(l_1^2 = 484 + (20 - x)^2\) and \(l_2^2 = 256 + x^2\).
\[ S(x) = l_1^2 + l_2^2 = [484 + (20 - x)^2] + [256 + x^2] \]
\[ S(x) = 484 + 400 - 40x + x^2 + 256 + x^2 \]
\[ S(x) = 2x^2 - 40x + 1140 \]
Step 2: Find the derivative and critical point
Differentiate \(S(x)\) with respect to \(x\):
\[ S'(x) = 4x - 40 \]
Set \(S'(x) = 0\) for the minimum value:
\[ 4x - 40 = 0 \implies 4x = 40 \implies x = 10 \]
Step 3: Verify minimum using second derivative
Find \(S''(x)\):
\[ S''(x) = 4 \]
Since \(S''(10) = 4 > 0\), the function has a local minimum at \(x = 10 \text{ m}\).
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Question: 4
If the \(22 \text{ m}\) long pole is also replaced by a \(16 \text{ m}\) long pole, at what distance from either pole should the ladders be kept so that the sum of squares of lengths of ladders needed to reach the top of the pole is minimum?
Show Hint
In symmetric optimization problems where the coefficients of the quadratic terms are the same, the minimum usually occurs exactly at the midpoint of the interval.
Concept: • When both poles are of equal height, the problem becomes symmetrical.
• Minimize the sum of squared lengths using the same differentiation procedure as before.
Step 1: Define the new function \(S_{new}(x)\)
Both heights are now \(16 \text{ m}\).
\[ l_1^2 = 16^2 + (20 - x)^2 = 256 + (20 - x)^2 \]
\[ l_2^2 = 16^2 + x^2 = 256 + x^2 \]
\[ S_{new}(x) = 256 + 400 - 40x + x^2 + 256 + x^2 \]
\[ S_{new}(x) = 2x^2 - 40x + 912 \]
Step 2: Find the critical point
\[ S'_{new}(x) = 4x - 40 \]
Setting the derivative to zero:
\[ 4x - 40 = 0 \implies x = 10 \]
Step 3: Interpret the result
The ladders should be placed at the midpoint of the road (\(10 \text{ m}\) from either pole) to minimize the sum of the squares of their lengths.
