MAXIMA AND MINIMA
Needless to say, this topic has been asked every time in board exams mostly for 6 marks else for 4 marks. The topic consists mainly of practical applications of derivatives and has a variety of questions surrounding it.
But before that, here are some basic terms and definitions.
CRITICAL POINTS- points where f’(c) =0 or f’(c) is not defined where c ∈ Domain f.
MAXIMUM/MINIMUM VALUE- let f be a function defined on interval I
f has a maximum value in I , if f(c) > f(x) ∀ x ∈ I.
f has a minimum value in I, if f(c) < f(x) ∀ x ∈ I.
HERE, f(c) is called maximum/minimum value of f in I and c is the point of maximum or minimum value in I.
MONOTONIC FUNCTION- A function f in interval I is said to be monotonic if and only if f is strictly increasing of strictly decreasing in I.
LOCAL MAXIMA/MINIMA- Let f be a real valued function and let c be an interior point in the domain of f.
c is called a point of LOCAL MAXIMA if there exists h>0 such that f(c)>f(x) ∀ x ∈ (c-h,c+h)
or a point of LOCAL MINIMA if there exists h>0 such that f(c)<f(x) ∀ x ∈ (c-h,c+h)
f(c) is then called the local maximum or local minimum value of f respectively.
ABSOLUTE MAXIMUM (MINIMUM)/GLOBAL MAXIMUM (MINIMUM)/GREATEST (LEAST) VALUE- These are the values of f in a closed interval I which are the largest(smallest) when domain of f is I.
Now, let us look at some of the tests we will use to check the points of MAXIMA AND MINIMA
FIRST DERIVATIVE TEST- Let f be a differentiable function in interval I.
STEP 1- Find critical points of f i.e. where f’(x) = 0 or f’(x) is not defined.
Now, select a small value of h> 0 , say 0.1 or 0.01.
STEP 2- let c be a critical point.
If f’(c-h)>0 and f’(c+h)<0, then c is a point of LOCAL MAXIMA.
OR
If f’(c-h)<0 and f’(c+h)>0, then c is a point of LOCAL MINIMA.
SECOND DERIVATIVE TEST- let f be a twice differentiable function.
STEP 1- Differentiate f and obtain critical points of f. Let the critical point(s) be c.
STEP 2- Differentiate f’ and obtain f’’.
If f’’(c)>0,
⇒c is a point of local minima.
f’’(c)<0,
⇒c is a point of local maxima.
If f’’(c)=0,
⇒SECOND DERIVATIVE TEST FAILS, use first derivative test.
Now that we are acquainted with theory, let us look at the type of questions asked.-
WORD PROBLEMS ON PRACTICAL APPLICATION-The most common question asked in Board examinations is that of practical application and is usually given as a 6 MARK QUESTION or sometimes as a 4 MARK QUESTION.
You will be given a word problem and you need to formulate the required mathematical expression and then apply the required tests.
CAUTION! Be careful to use the details given in the question. Use only those details which are given in the question and not the ones which need to be proved.
FOLLOWING are some sample questions with their solution-
QUESTION 1. An open box which has a square base , is to be made out of a given quantity of metal sheet of area C2. Prove that the maximum volume of the box is C3/6√3.
ANSWER :
Let the side of the square base be x and height of the open box be y. Then C2 = Surface area of the open box = x2 + 4 xy Or, y = (C2 – x2)/4x.
Volume of the box, V = x2.y = x2. (C2 – x2)/4x Or, V = 1/4 (C2x – x2) -- (1) Therefore, dV/dx = 1/4 (C2 – 3x2)
d²V/dx² = – 3/2 x -- (2) For V to be maximum or minimum, dV/dx = 0 Or, C2 – 3x2 = 0
=> x = ± C/√3. x can’t be negative (x being length)
so x = C/√3. Therefore, d²V/dx² x = C/√3 = – 3/2. C/√3 – (√3/2)C < 0 [ As, C > 0] Therefore, V is maximum. Maximum value of V = 1/4[C2. C/√3 – (C/√3)3] = C3/6√3.
HENCE PROVED.
QUESTION 2. A right-angled triangle with constant area S is given. Prove that the hypotenuse of the triangle is least when the triangle is isosceles.
ANSWER:
Let length of the hypotenuse be l of the given right-angled triangle ABC at B and LCAB = θ, 0 < θ < π/2. AB = l cos θ and BC = l sin θ. Area of Δ ABC = 1/2 (l cos θ)(l sin θ) = S (given). Or, l2/4 sin 2θ = S Or, l2 = 4S cosec 2θ Writing l2 as f(θ) we get, f(θ) = 4S cosec 2θ ---------------- (i) Differentiating (i) w.r.t. θ we get,
f’(θ) = 4S (– cosec 2θ cot 2θ).2 = – 8 S cosec 2θ cot 2θ and f”(θ) = – 8S{cosec 2 θ (– cosec2 2θ).2 + cot 2θ(– cosec 2θcot 2θ).2} = 16S cosec 2θ (cosec2 2θ + cot2 2θ). Now, f’(θ) = 0
=> – 8 S cosec 2θ cot 2θ = 0 => (1/sin 2θ)(cos 2θ/sin 2θ) = 0 => sin 2θ = 0 => 2θ = π/2 => θ = π/4. Also, f”(π/4) = 16 S cosec π/2 (cosec2 π/2 + cot2 π/2) = 16.1.(1 + 0) = 16 S > 0 Therefore, f(θ) is least when θ = π/4 Or, l is least when θ = π/4. When θ = π/4, AB = l cos π/4 = l/√2 and BC = l sin π/4 = l/√2. Hence, the hypotenuse is the least when AB = BC i.e. the triangle is isosceles.
QUESTION 3- AB is a diameter of the circle and C is any point on the circle. Show that the area of the triangle ABC is maximum when the triangle is isosceles.
ANSWER:
Ans- since angle C is made by a semicircle ,
∴ ∠C =90
AB²= AC²+BC² (Pythagoras theroem) Let AB=c , AC= x and BC= y ⇒c²=x²+y² - (1) let area of triangle be denoted by A. A = xy/2 A²=x²y²/4=B Now, A will be maximised when B is maximised B=x²(c²-x²)/4 ⇒dB/dx=x²(-2x)/4 + 2x(c²-x²)/4 for critical points , dB/dx=0 ⇒-4x³ + 2xc²=0 ⇒c²=2x² ( x≠ 0) ⇒ c= √2x on substituting the value of c in equation 1, we have x = y = c/√2 d²B/dx²= -3x² + c²/2 = -2x²<0 ⇒ area is minimum when x = y = c/√2 or when triangle is isosceles.
FINDING ABSOLUTE MAXIMUM/MINIMUM- in this type of question, you will be provided with a function over an interval I and you need to find its maximum and minimum value.
Following is a sample question on the aforementioned type-
QUESTION 1- Find the maximum and minimum values of f(x) = sin²(x) + cos(x) in [0, ∏/2]
f(x) = sin²(x) + cos(x) f’(x) = 2sin(x)cos(x) - sin(x)s for obtaining critical points , put f’(x) = 0 ⇒2sin(x)cos(x)=sin(x) ⇒sin(x)[2cos(x) - 1]=0 ⇒sin(x)=0 or cos(x) = 1/2 ⇒ x = 0 or x = ∏/3 also, we require end points of interval, i.e. x=0 , ∏/2 f(0) = 1 f(∏/3) = 3/4 + 1/2 = 5/4 f(∏/2) = 1 hence, absolute maximum value is 5/4 and occurs at x =∏/3. Absolute minimum value is 1 and occurs at x=0 , ∏/2.
QUESTION 2 - Find the absolute maximum and minimum of function f(u) = 2t³ + 3t² - 12t -7 in [-1,3].
f(u) = 2t³ + 3t² - 12t -7 f’(u) = 6t² + 6t - 12 for critical points , f’(u) = 0 ⇒6(t²+t-2)=0 ⇒ 6(t+2)(t-1)=0 ⇒ t= -2 or t = 1 since t = -2 ∉[-1,3] ⇒ t= 1 is the only critical point. end points of interval are -1,3 f(-1) = 6 f(1) = -14 f(3)= 38 ⇒ Absolute minima is -14 and occurs at x= 1 Absolute maxima is 38 and occurs at x= 3
SOME UNSOLVED QUESTIONS FOR PRACTICE.
Question 1- Prove that a triangle with given two sides has its area maximum when the angle between them is π/2.
QUESTION 2- An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width.
QUESTION 3- Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6√3r.
QUESTION 4 - Find absolute maxima and absolute minima of the following functions in the given interval.
i) f(y) = sin(y/3) + 2y/9 , y∈ [-10,15]
ii) f(x)= ln(x²+2x+14) , x∈[-4,2]
Comments