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CBSE Sample Paper 2023-24 · Section B · 2 marks

Q23: Maximum value of 14x2+2x+1\frac{1}{4x^2 + 2x + 1}⁠, or the maximum profit P(x)=72+42x−x2P(x) = 72 + 42x - x^2

Sample 2023-24Ch 6 · Application of Derivatives
The question

(a) If f(x)=14x2+2x+1f(x) = \dfrac{1}{4x^2 + 2x + 1}⁠; x∈Rx \in \mathbb{R}⁠, then find the maximum value of f(x)f(x)⁠.

OR

(b) Find the maximum profit that a company can make, if the profit function is given by P(x)=72+42x−x2P(x) = 72 + 42x - x^2⁠, where xx is the number of units and PP is the profit in rupees.

Exam answer

(a)

Let g(x)=4x2+2x+1g(x) = 4x^2 + 2x + 1⁠. Then

Why?

f(x)=1g(x)f(x) = \frac{1}{g(x)} is largest when its denominator g(x)g(x) is smallest, as long as g(x)g(x) stays positive. So we look for the least value of g(x)g(x)⁠.

g(x)=4(x2+2x⋅14+116)+34=4(x+14)2+34≥34\begin{aligned} g(x) &= 4\left(x^2 + 2x\cdot\frac14 + \frac{1}{16}\right) + \frac34 \\ &= 4\left(x + \frac14\right)^2 + \frac34 \\ &\ge \frac34 \end{aligned}
Why?

This is completing the square. Take 4 out of the xx terms:

4x2+2x=4(x2+x2)4x^2 + 2x = 4\left(x^2 + \frac{x}{2}\right)

(Inside the bracket, 2x4=x2\frac{2x}{4} = \frac{x}{2}⁠.)

Half of the coefficient 12\frac12 is 14\frac14⁠, and (14)2=116\left(\frac14\right)^2 = \frac{1}{16}⁠. Adding 116\frac{1}{16} inside the bracket adds 4×116=144 \times \frac{1}{16} = \frac14⁠, so the constant left outside is 1−14=341 - \frac14 = \frac34⁠.

g(x)=4(x2+x2)+1=4(x2+x2+116)−4×116+1=4(x+14)2−14+1=4(x+14)2+34\begin{aligned} g(x) &= 4\left(x^2 + \frac{x}{2}\right) + 1 \\ &= 4\left(x^2 + \frac{x}{2} + \frac{1}{16}\right) \\ &\quad - 4 \times \frac{1}{16} + 1 \\ &= 4\left(x + \frac14\right)^2 - \frac14 + 1 \\ &= 4\left(x + \frac14\right)^2 + \frac34 \end{aligned}

The bracket is a perfect square because (x+14)2=x2+2x⋅14+116\left(x + \frac14\right)^2 = x^2 + 2x\cdot\frac14 + \frac{1}{16}⁠, and 2x⋅14=x22x \cdot \frac14 = \frac{x}{2}⁠. Also −14+1=−14+44=34-\frac14 + 1 = -\frac14 + \frac44 = \frac34⁠.

A square is never negative, so 4(x+14)2≥04\left(x + \frac14\right)^2 \ge 0⁠, which gives the last line.

So g(x)≥34>0g(x) \ge \frac34 > 0⁠, with g(x)=34g(x) = \frac34 at x=−14x = -\frac14⁠. Hence f(x)=1g(x)≤43f(x) = \frac{1}{g(x)} \le \frac43⁠.

Why?

The square is 0 exactly when x+14=0x + \frac14 = 0⁠, that is at x=−14x = -\frac14⁠. There gg takes its least value 34\frac34⁠.

For positive numbers, taking reciprocals reverses an inequality:

g(x)≥34⇒1g(x)≤43g(x) \ge \frac34 \Rightarrow \frac{1}{g(x)} \le \frac43

Here 134=1×43=43\frac{1}{\frac34} = 1 \times \frac43 = \frac43 (to divide by a fraction, multiply by its flip).

Watch out: the maximum of ff comes from the minimum of gg⁠, not from the maximum of gg⁠.

∴\therefore Maximum value of f(x)=43f(x) = \frac43 (at x=−14x = -\frac14⁠).

(b)

P(x)=72+42x−x2P(x) = 72 + 42x - x^2

For maxima and minima,

P′(x)=0⇒42−2x=0P'(x) = 0 \Rightarrow 42 - 2x = 0
Why?

At a maximum or a minimum the tangent is flat, so the derivative is 0.

Differentiate term by term: the constant 72 gives 0, 42x42x gives 42, and x2x^2 gives 2x2x⁠. So P′(x)=42−2xP'(x) = 42 - 2x⁠.

⇒x=21 and P′′(x)=−2<0\Rightarrow x = 21 \text{ and } P''(x) = -2 < 0
Why?

42−2x=042 - 2x = 0 gives 2x=422x = 42⁠, so x=21x = 21⁠.

42−2x=042=2x(add 2x)x=422=21\begin{aligned} 42 - 2x &= 0 \\ 42 &= 2x \quad (\text{add } 2x) \\ x &= \frac{42}{2} = 21 \end{aligned}

Second derivative test: differentiate 42−2x42 - 2x once more to get −2-2⁠.

P′′(c)<0⇒maximumP′′(c)>0⇒minimum\begin{aligned} P''(c) &< 0 \Rightarrow \text{maximum} \\ P''(c) &> 0 \Rightarrow \text{minimum} \end{aligned}

So, P(x)P(x) is maximum at x=21x = 21⁠.

P(21)=72+42×21−212=72+882−441=513\begin{aligned} P(21) &= 72 + 42 \times 21 - 21^2 \\ &= 72 + 882 - 441 = 513 \end{aligned}
Why?

42×21=88242 \times 21 = 882 and 212=44121^2 = 441⁠, so 72+882−441=51372 + 882 - 441 = 513⁠.

42×21=840+42=882212=21×21=44172+882=954954−441=513\begin{aligned} 42 \times 21 &= 840 + 42 = 882 \\ 21^2 &= 21 \times 21 = 441 \\[4pt] 72 + 882 &= 954 \\ 954 - 441 &= 513 \end{aligned}

Watch out: the question asks for the maximum profit, which is P(21)P(21)⁠, not the number of units x=21x = 21⁠.

∴\therefore The maximum profit is ₹ 513.

Where this comes from

Chapter 6, Application of Derivatives: Maximum value of a function; maximum profit.

Question type: (a) Maximum and minimum values you can see without calculus; (b) Local maxima and minima

How to spot it: click to see

(a) Type 6. Maximum and minimum values you can see without calculus

Spot it: a square plus a number, a modulus plus a number, or sin or cos plus a number.

Here: Completing the square gives 4x2+2x+1=4(x+14)2+34≥344x^2 + 2x + 1 = 4\left(x + \frac14\right)^2 + \frac34 \ge \frac34⁠, least at x=−14x = -\frac14⁠. So ff⁠, its reciprocal, is at most 43\frac43⁠.

(b) Type 7. Local maxima and minima: the second derivative test

Spot it: "Find the local maxima and local minima" or "the local maximum value".

Here: The profit function is given: P′(x)=42−2x=0P'(x) = 42 - 2x = 0 at x=21x = 21⁠, and P′′(x)=−2P''(x) = -2 is negative, so x=21x = 21 gives the maximum, P(21)=513P(21) = 513⁠.

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