 # Maxima and Minima

• A point c in the domain of a function ƒ at which either ƒ'(c)=0 or ƒ is not differentiable is called a critical point of ƒ.
• First Derivative Test Let ƒ be a function defined on an open interval I. Let ƒ be continuous at a critical point c in I. Then
(i) If ƒ'(x) changes sign from positive to negative as x increases through c, i.e., if ƒ'(x) > 0 at every point sufficiently close to and to the left of c, and ƒ'(x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If ƒ'(x) changes sign from negative to positive as x increases through c, i.e., if ƒ'(x) < 0 at every point sufficiently close to and to the left of c, and ƒ'(x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If ƒ'(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflexion.
• Second Derivative Test Let ƒ be a function defined on an interval I and c ∈ I. Let ƒ be twice differentiable at c. Then
(i) x = c is a point of local maxima if ƒ'(c) = 0 and ƒ''(c) < 0
The values of ƒ(c) is local maxima value of ƒ.
(ii) x = c is a point of local minima if ƒ'(c) = 0 and ƒ''(c) > 0
In this case, ƒ(c) is local minimum value of ƒ.
(iii) The test fails if ƒ'(c)=0 and ƒ''(c) = 0.
In this case, we go back to the first derivative test and find whether c is a point of maxima, minima or a point of inflexion.
• Working rule for finding absolute maxima and/or absolute minima
Step 1: First all critical points of ƒ in the interval, i.e., find points x where either ƒ'(x) = 0 or ƒ is not differentiable.
Step 2: Take the end points of the interval.
Step 3: At all these points (listed in Step 1 and 2), calculate the values of ƒ.
Step 4: Identify the maximum and minimum values of ƒ out of the vales calculated in Step 3. This maximum value will be the absolute maximum value of ƒ and the minimum value will be the absolute minimum value of ƒ.

### Part3: View the Topic in this video From 00:40 To 51:44

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Tricks on Maxima and Minima:

• If f(x) = x2 + k, where k is real numbers then f(x) ≥ k ∀ x ∈ R, then minimum value of f(x) is k.
• If f(x) = x2 + k, where k is real number and f(x) ≤ k ∀ x ∈ R, then the maximum value of f(x) is k.
• The maximum value of f(x) = a cos2x + b sin2x is a and minimum value is b. (if a > b)
• The minimum value of f(x) = a tan x + b cot x is 2\sqrt{ab} and attains at x=\tan^{-1}\sqrt{\frac{b}{a}}.
• The minimum value of f(x) = a2 sec2x + b2 cosec2 x is (a + b)2 and attains at x=\tan^{-1}\sqrt{\frac{b}{a}}
• The minimum value of f(x) = a sec x + b cosec x is (a2/3 + b2/3)3/2 and attains at x=\tan^{-1}\left(\frac{a}{b}\right)^{1/3}
• The maximum value of f(x) = sinmx . cosnx is \frac{m^{\frac{m}{2}}\cdot n^{\frac{n}{2}}}{(m+n)^{\frac{m+n}{2}}} and attains at x=\tan^{-1}\left(\sqrt{\frac{m}{n}}\right)
• ax2 + bx + c > 0 ∀ x ∈ R ⇒ a > 0 and Δ < 0.
• ax2 + bx + c < 0 ∀ x ∈ R ⇒ a < 0 and Δ < 0.
• The sum of two numbers is k. If the sum of their squares is minimum, then the numbers are \frac{k}{2},\frac{k}{2}
• The sum of two numbers is k. If the product is maximum, then the numbers are \frac{k}{2},\frac{k}{2}
• The product of two numbers is k. If the sum of their squares is minimum, then the numbers are \sqrt{k},\sqrt{k}
• The sum of two numbers is k. If the product of the square of first and cube of the second is maximum, then the numbers are \frac{2k}{5},\frac{3k}{5}.
• If a > 0, b > 0, x > 0, the least value of f(x)=ax+\frac{b}{x} is 2\sqrt{ab}.
• The maximum area of rectangle inscribed in a circle of radius r is 2r2.
• The maximum area of the triangle inscribed in the circle of radius r is \frac{3\sqrt{3}}{4} \ r^{2}.
• The maximum area of a isosceles triangle inscribed in a ellipse whose one vertex is coincides with one end of major axis is \frac{3\sqrt{3}}{4} \ ab.
• The perimetre of sector is c. Then the maximum area of sector is \frac{c^{2}}{16}.
• The area of sector is c. Then the least perimeter of sector is 4\sqrt{c}.
• The hypotenuse of right angle triangle is a. If the area of the triangle is maximum. Then the sides are \frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}.
• If the hypotenuse of right angled triangle is at a distance of a and b from the sides, then the minimum length of hypotenuse is (a2/3 + b2/3)3/2.
• The area of a greatest rectangle inscribed in the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 is 2ab.
• Two sides of a triangle are given. The area of the triangle is maximum, then the angle between the sides is \frac{\pi}{2}.
• The sum of hypotenuse and one side of right angled triangle is given. The area is maximum then the angle between them is \frac{\pi}{3}.
• An open box of maximum volume is made from a rectangular piece of tin of length 'a' and bredth 'b' by culting four equal square pieces from four corners and folding up the tin. Then the length of the box is \frac{(a+b)-\sqrt{a^{2}+b^{2}-ab}}{6}
• A cone of maximum volume that can be inscribed in the sphere of radius (r) is of height is \frac{4r}{3}
• A right circular cylinder of maximum volume that can be inscribed in sphere of radius r is of height \frac{2r}{\sqrt{3}}.
• If x + y = a and xm − yn is maximum then x and y are \frac{am}{m+n},\frac{an}{m+n}
• If mx + ny = k, then xy is maximum x : y = n : m.
• If f(x) = (x − a)(x − (a + d)) (x − (a + 2d)) then local maximum and minimum occurs at x=(a+d)-\frac{d}{\sqrt{3}},x=(a+d)+\frac{d}{\sqrt{3}} and maximum and minimum values are \frac{2d}{3\sqrt{3}},\frac{-2d}{3\sqrt{3}}.