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Maxima and minima of a function can be found by using the concept of derivatives. Derivatives are used to find the gradient/ slope of the function, the points where the gradient is zero are called turning points/stationary points. These points give the largest or smallest values (locally) of the function.
Read More: Application of Derivatives
Key Terms: Intervals, Derivatives, function, Maxima and Minima, First Order Derivative Test, Second Derivative Test.
Maximum
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Let, f(x) is the real function having an I interval. Then, f(x) will have the maximum value in the interval I if and only if a point “c” in the interval satisfies f(x) ≤ f(c) for all values of x ? I.
So, f(c) will be the maximum value function f(x) in an I interval, and the point c will be known as the maximum value point of the f function in Interval I.
Minimum
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Let, f(x) is a real function having an interval I. Then f(x) will have the minimum value in the interval I if there is a point “c” in the interval satisfy f(x) ≥ f(c) for all values of x ? I.
So, f(c) will be the minimum value of f(x) in the given interval I, and point c will be known as the minimum value in the given interval I.
f(x) an extreme value function if a point c in the given interval I such that f(c) is either a maximum value or a minimum value of f(x) in the interval I and the point c is known as an extreme point.
Read More: Continuity and Differentiability
Local Maxima
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Let, f(x) function have a local maxima at the point x = a, if a neighbourhood point such as (a-δ, a+δ) satisfy the condition like
f(x) < f(a) for all the values of x (a-δ, a+δ), x ≠ a.
OR, f(x) – f(a) < 0 for all the values of x (a – δ, a + δ), x ≠ a.
So, the local maxima of f(x) at x = a will be f(a).
Read More: Area of a Triangle
Local Minima
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Let, f(x) function have a local minima at the point x = a, if a neighbourhood point such as (a-δ, a+δ) satisfy the condition like
f(x) > f(a) for all the values of x (a-δ, a+δ), x≠a.
OR, f(x) – f(a) > 0 for all the values of x (a-δ, a+δ), x≠a.
So, the local minima of f(x) at x = a will be f(a).
Read More: Increasing and decreasing functions
Derivative Tests
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The derivative test is the easiest way to find the maxima and minima of any function. The first-order derivative and second-order derivative of a function are used in the Derivative test.
- First Order Derivative Test
Let c is the continuous function defined in an open interval I having a critical point c in I such that f’(c) = 0.
- If there is a sign change in f’(x) from positive to negative then x increases through point c, which is known as local maxima and the f(c) is the maximum value.
- If there is a sign change in f’(x) from negative to positive then x increases through point c, which is known as local minima and the f(c) is the minimum value.
- If the sign of f’(x) doesn’t change then x increases through c and neither having local nor a point of local maxima. Then it is known as point of inflection.
- Second Derivative Test
Let f be the two times differentiable function at c in an interval I.
- If x = c is the point of local maxima if f'(c) = 0 and f”(c)<0, then f(c) will have local maximum value.
- If x = c is the point of local minima if f'(c) = 0 and f”(c) < 0, then f(c) will have local minimum value.
- When both f'(c) = 0 and f”(c) = 0, then the test fails and the first derivative test will give the value of local maxima and minima.
Read Also: Tangents and normals
Calculation of Maximum and Minimum Values in a Closed Interval
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If f(x) be a continuous function on an interval I = [a, b], then f(x) has the absolute maximum and absolute minimum value if f(x) attains it at least once in interval I.
Let f(x) be a differentiable closed function and c be any interior point of the interval I.
Then f′(c) = 0 if f(x) attains its absolute maximum and absolute minimum value at c.
Example-
- Calculate the absolute maximum and minimum values of f(x) = 2x3 – 15x2 + 36x + 1 in the closed interval of [1, 5]?
Solution: f(x) = 2x3 – 15x2 + 36x + 1
Now, f′(x) = 6x2 – 30x + 36 = 6(x – 3) (x – 2)
For, f′(x) = 0 the solution of x will be x = 2 and x = 3.
Calculate the value of f(x) at these points and at the end points of the interval [1, 5].
For, x = 1, f(1) =2(13) – 15(12) + 36(1) + 1 = 24
For, x = 2, f(2) =2(23) – 15(22) + 36(2) + 1 = 29
For, x = 3, f(3) =2(33) – 15(32) + 36(3) + 1 = 28
For, x = 5, f(5) =2(53) – 15(52) + 36(5) + 1 = 56
So, the absolute maximum value off on [1, 5] is 56, at x =5, and absolute minimum value is 24 on [1, 5] at x = 1.
- Calculate the values of two positive numbers whose sum is 15 and the sum of whose squares is minimum?
Solution: Let one of the numbers be x. Then the other number is (15 – x).
Let, the sum of the squares of these numbers is denoted by S(x).
So, S(x) = x2 + (15 – x)2 = 2x2 – 30x + 225
S’ (x)= 4x- 30 and S”(x)= 4
Now S′(x) = 0 gives x=15/2. Also S” (15/2)= 4> 0.
Second derivative test proves that x=15/2 is the point of local minima of S.
Hence the two numbers are 15/2 and 15- 15/2= 15/2.
- Find the maxima and minima for f(x) = 2x3 – 21x2 + 36x – 15
Solution: We have f(x) = 2x3 – 21x2 + 36x – 15
f’(x) = 6x2 – 42x + 36
For, f’(x) = 0 ⇒ 6x2 – 42x + 36 = 0
⇒ x2 – 7x + 6 = 0
⇒ (x-6) (x-1) = 0
⇒ x = 6 or x = 1 are the points where minima or maxima will be observed.
For, f’’(x) = 12x – 42
At the point, x = 1, f”(1) = 12 – 42 = – 30 <0
So, at x = 1, local maximum is observed.
The maximum value is f(1) = 2 – 21 + 36 – 15 = 2
At x= 6, f’’(x) = 12 × 6 – 42 = 30 > 0
So, the local minimum can be found at x = 6.
The local minimum value is f (6) = 2(6)3 – 21(6)2 + 36(6) – 15
= 2 × 216 – 21 × 36 + 216 – 15
= 432 – 756 + 216 – 15 = – 123
Also Read:
Things to Remember
- Maxima and minima of a function can be found by using the concept of derivatives.
- Derivatives are used to find the gradient/ slope of the function, the points where the gradient is zero are called turning points/stationary points.
- The derivative test is the easiest way to find the maxima and minima of any function.
- The first-order derivative and second-order derivative of a function are used in the Derivative test.
Sample Questions
Ques:How to calculate local minima from second derivative test? (1 mark)
Ans. If x = c is the point of local minima if f'(c) = 0 and f”(c) < 0, then f(c) will have local minimum value.
Ques: How to calculate maxima and minima at a close interval? (1 mark)
Ans. If f(x) be a continuous function on an interval I = [a, b], then it has the absolute maximum and absolute minimum value if f(x) attains at least one values in interval I. Then calculate the values of f(x) by putting a, b and absolute values.
Ques: What is the critical point? (1 mark)
Ans: A point c in the domain of a function at which either f′(c) = 0 or the function is not differentiable is called a critical point of the function.
Ques: A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs. 70 per square meter for the base and Rs. 45 per square meter for the sides, what is the cost of the least expensive tank? (2019)
Ans: Let’s take length, breadth and height be l, b and h of the tank, respectively.
h = 2m
Volume (v) = 8m3
v = l*b*h
Therefore l*b*2 = 8
Lb = 4
B = 4/l
We can say area of the base (lb) = 4m2
The area of the 4 walls of the tank (A) = 2h(l+b)
Therefore A = 4(l + 4/l)
dA/dl = 4(1 - 4/l2)
For A’s maximum and minimum values, we must have dA/dl = 0
4(1 - 4/l2) = 0
L = +-2
Since we know the length can’t be negative.
Hence, l = 2m
Therefore b = 4/2 = 2m
Then,
d2A/dl2 = 32/l3
At l =2
d2A/dl2 = 32/8 = 4>0
Therefore the area can be considered the minimum when l = 2m
We know l = b = h = 2m
The cost of building the base = Rs 70 * (lb) = Rs 70 * 4 = Rs 280
The cost of building the walls - Rs 2h (l+b) * 45 = Rs90 (2) (2+2) = Rs 8(90) = Rs 720
Total cost incurred for the tank = Rs (280 + 720) = Rs 1000
Ques. Find the point on the curve y2 = 4x, which is nearest to the point (2, – 8) (2019 outside Delhi)
Ans: Point (x1, y1) on the curve
y12 = 4x1
D is always considered positive
D2 = (x1-2)2 + (y1 +8)2
= (y12 /4 - 2)2 + (y1 + 8)2
dD2/dy1 = 2(y12/4 - 2)2y1 /4 + 2(y1 +8)
= y1(y12 -8) / 4 +2y1 + 16
= y13 - 8y1 + 8y1 + 64
= y13 + 64
dD2/dy1 = 0 y13 = -64
y1 = – 4
(y1)2 = 4x1
(-4)2 = 4x1
x1 = 4
Point is (4, -4)
Ques. 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 material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question? (2018)
Ans: The tank mentioned in the question has a square base and let the units x, x, y be the length, width and height respectively.
Volume = length*breadth*height = x2y
Total area of the surface = 2(lb + bh + hl) - lb = x2 + 4x
The volume of the tank is said to be constant
Then the surface area = x2 + 4xy
For the least total surface area
Therefore when x = 2y the surface area can be considered to be the least i.e., the width of the tank is double its depth.
Hence if the surface area of the sheet is minimum the price of the sheet will be the least as well making it both cost-effective and economical.
Ques. If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is π/3. (2017)
Ans:
Ques. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3. Also find maximum volume in terms of volume of the sphere. (2016)
Ans:
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