Maharashtra Board Class 12 Maths Part 1 Chapter 4 Applications of Derivatives 4.3 Solutions

Question 1. Determine the maximum and minimum values of the following functions:
(i) f(x) = \(2x^3 - 21x^2 + 36x - 20\)
Answer: Solution:
f(x) = \(2x^3 - 21x^2 + 36x - 20\)
∴ f'(x) = \(\frac{d}{dx}(2x^3 - 21x^2 + 36x - 20)\)
= \(2 \times 3x^2 - 21 \times 2x + 36 \times 1 - 0\)
= \(6x^2 - 42x + 36\)
and f"(x) = \(\frac{d}{dx}(6x^2 - 42x + 36)\)
= \(6 \times 2x - 42 \times 1 + 0\)
= \(12x - 42\)
f'(x) = \(0\) gives \(6x^2 - 42x + 36 = 0\).
∴ \(x^2 - 7x + 6 = 0\)
∴ \((x - 1)(x - 6) = 0\)
∴ the roots of f'(x) = \(0\) are \(x_1 = 1\) and \(x_2 = 6\).
For \(x = 1\), f"(1) = \(12(1) - 42 = -30 < 0\)
∴ by the second derivative test,
f has maximum at \(x = 1\) and maximum value of f at \(x = 1\)
f(1) = \(2(1)^3 - 21(1)^2 + 36(1) - 20\)
= \(2 - 21 + 36 - 20\)
= \(-3\)
For \(x = 6\), f"(6) = \(12(6) - 42 = 30 > 0\)
∴ by the second derivative test,
f has minimum at \(x = 6\) and minimum value of f at \(x = 6\)
f(6) = \(2(6)^3 - 21(6)^2 + 36(6) - 20\)
= \(432 - 756 + 216 - 20\)
= \(-128\)
Hence, the function f has maximum value \(-3\) at \(x = 1\) and minimum value \(-128\) at \(x = 6\). In simple words: To find maximum/minimum values, we first find the first derivative f'(x) and set it to zero to find critical points. Then, we use the second derivative f''(x) to check if these points yield a maximum (f'' < 0) or a minimum (f'' > 0), and calculate the function value at these points.

🎯 Exam Tip: Remember to clearly state the maximum/minimum values along with the corresponding x-values. Showing all steps for derivatives and critical points is crucial for scoring.

(ii) f(x) = \(x \cdot \log x\)
Answer: Solution:
f(x) = \(x \cdot \log x\)
f'(x) = \(\frac{d}{dx}(x \cdot \log x)\)
= \(x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x)\)
= \(x \times \frac{1}{x} + (\log x) \times 1\)
= \(1 + \log x\)
and f"(x) = \(\frac{d}{dx}(1 + \log x)\)
= \(0 + \frac{1}{x}\)
= \(\frac{1}{x}\)
Now, f'(x) = \(0\), if \(1 + \log x = 0\)
i.e. if \(\log x = -1 = -\log e\)
i.e. if \(\log x = \log(e^{-1}) = \log \frac{1}{e}\)
i.e. if \(x = \frac{1}{e}\)
When \(x = \frac{1}{e}\), f"(\frac{1}{e}) = \(\frac{1}{(1/e)} = e > 0\)
∴ by the second derivative test,
f is minimum at \(x = \frac{1}{e}\)
Minimum value of f at \(x = \frac{1}{e}\)
= \(\frac{1}{e} \log(\frac{1}{e})\)
= \(\frac{1}{e} \log(e^{-1})\)
= \(\frac{1}{e}(-1) \log e\)
= \(-\frac{1}{e}\) [......... \(\log e = 1\)]
Hence, the function f has minimum at \(x = \frac{1}{e}\) and minimum value is \(-\frac{1}{e}\). In simple words: For functions involving logarithms, apply the product rule for differentiation and then the second derivative test to determine the nature (minimum/maximum) and value of the function at critical points.

🎯 Exam Tip: Be careful with logarithmic properties, especially when solving for critical points. Ensure correct application of derivative rules for log functions.

(iii) f(x) = \(x^2 + \frac{16}{x}\)
Answer: Solution:
f(x) = \(x^2 + \frac{16}{x}\)
∴ f'(x) = \(\frac{d}{dx}(x^2 + \frac{16}{x})\)
= \(2x + 16(-1)x^{-2} = 2x - \frac{16}{x^2}\)
and f"(x) = \(\frac{d}{dx}(2x - \frac{16}{x^2})\)
= \(2 \times 1 - 16(-2)x^{-3} = 2 + \frac{32}{x^3}\)
f'(x) = \(0\) gives \(2x - \frac{16}{x^2} = 0\)
∴ \(2x^3 - 16 = 0\)
∴ \(x^3 = 8\)
∴ \(x = 2\)
For \(x = 2\), f"(2) = \(2 + \frac{32}{(2)^3} = 2 + \frac{32}{8} = 2 + 4 = 6 > 0\)
∴ by the second derivative test, f has minimum at \(x = 2\) and minimum value of f at \(x = 2\)
f(2) = \((2)^2 + \frac{16}{2}\)
= \(4 + 8\)
= \(12\)
Hence, the function f has a minimum at \(x = 2\) and a minimum value is \(12\). In simple words: To find the minimum for a function like this, differentiate twice. Set the first derivative to zero to find the critical point(s). Use the second derivative to confirm it's a minimum (positive value) and then calculate the function's value at that point.

🎯 Exam Tip: Pay attention to negative exponents when differentiating. A common error is incorrectly handling the derivative of \(1/x^n\). Double-check arithmetic, especially with fractions.

Question 2. Divide the number 20 into two parts such that their product is maximum.
Answer: Solution:
Let the first part of 20 be x.
Then the second part is \(20 - x\).
∴ their product = \(x(20 - x) = 20x - x^2 = f(x)\) .....(Say)
∴ f'(x) = \(\frac{d}{dx}(20x - x^2)\)
= \(20 \times 1 - 2x\)
= \(20 - 2x\)
and f"(x) = \(\frac{d}{dx}(20 - 2x)\)
= \(0 - 2 \times 1\)
= \(-2\)
The root of the equation f'(x) = \(0\)
i.e. \(20 - 2x = 0\) is \(x = 10\)
and f"(10) = \(-2 < 0\)
∴ by the second derivative test, f is maximum at \(x = 10\).
Hence, the required parts of 20 are 10 and 10. In simple words: To maximize the product of two parts of a number, represent one part as 'x' and the other as 'total-x'. Form a product function, differentiate it, set to zero to find 'x', and use the second derivative to confirm it's a maximum.

🎯 Exam Tip: Problems involving maximizing/minimizing quantities often require setting up a function based on the given conditions, then applying differentiation. Clearly define variables and the objective function.

Question 3. A metal wire of 36 cm long is bent to form a rectangle. Find its dimensions where its area is maximum.
Answer: Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Then its perimeter is \(2(x + y) = 36\)
∴ \(x + y = 18\)
∴ \(y = 18 - x\)
Area of the rectangle = \(xy = x(18 - x)\)
Let f(x) = \(x(18 - x) = 18x - x^2\)
Then f'(x) = \(\frac{d}{dx}(18x - x^2)\)
= \(18 \times 1 - 2x\)
= \(18 - 2x\)
and f"(x) = \(\frac{d}{dx}(18 - 2x)\)
= \(0 - 2 \times 1\)
= \(-2\)
Now, f(x) = \(0\), if \(18 - 2x = 0\)
i.e. if \(x = 9\)
and f"(9) = \(-2 < 0\)
∴ by the second derivative test, f has maximum value at \(x = 9\)
When \(x = 9\), \(y = 18 - 9 = 9\)
Hence, the rectangle is a square of side 9 cm. In simple words: To maximize the area of a rectangle with a fixed perimeter, express one dimension in terms of the other using the perimeter. Form an area function, differentiate, set to zero, and confirm it's a maximum using the second derivative. This shows a square has the maximum area for a given perimeter.

🎯 Exam Tip: Optimization problems require translating word problems into mathematical functions. Always derive both dimensions (length and breadth) to fully answer the question, especially if they are different variables.

Question 4. The total cost of producing x units is \((x^2 + 60x + 50)\) and the price is \((180 - x)\) per unit. For what units is the profit maximum?
Answer: Solution:
Let the number of units sold be x.
Then profit = S.P. - C.P.
∴ P(x) = \((180 - x)x - (x^2 + 60x + 50)\)
∴ P(x) = \(180x - x^2 - x^2 - 60x - 50\)
∴ P(x) = \(120x - 2x^2 - 50\)
P'(x) = \(\frac{d}{dx}(120x - 2x^2 - 50)\)
= \(120 \times 1 - 2 \times 2x - 0\)
= \(120 - 4x\)
and P"(x) = \(\frac{d}{dx}(120 - 4x)\)
= \(0 - 4 \times 1\)
= \(-4\)
P'(x) = \(0\) if \(120 - 4x = 0\)
i.e. if \(x = 30\) and P"(30) = \(-4 < 0\)
∴ by the second derivative test, P(x) is maximum when \(x = 30\).
Hence, the number of units sold for maximum profit is 30. In simple words: To find maximum profit, define the profit function as (Total Revenue - Total Cost). Differentiate this function with respect to the number of units (x), set the derivative to zero to find the optimal 'x', and use the second derivative to confirm it yields a maximum profit.

🎯 Exam Tip: Clearly distinguish between cost, revenue, and profit functions. Remember that total revenue is (price per unit) × (number of units). Always check the second derivative to confirm maximum or minimum.