Maxima and Minima

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Maxima and minima are known as the extrema of a function. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges. For the function, under the entire range, the maximum value of the function is known as the global(absolute ) maxima and the minimum value is known as the global (absolute) minima.
There are other maxima and minima of a function, which are not the absolute maxima and minima of the function and are known as local maxima and local minima

Maxima र minima फलनको छेउ-छेउको मान भनिन्छ। Maxima र minima भनेको फलनको दायराहरूबिचमा हुने अधिकतम वा न्यूनतम मान हो। फलनको सम्पूर्ण दायराहरूबिचमा हुने अधिकतम मानलाई global( absolute ) maxima भनिन्छ र सम्पूर्ण दायराहरूबिचमा हुने न्यूनतम मानलाई विश्वव्यापी global( absolute ) minima भनिन्छ।
यस बाहेक फलनक अन्य अधिकतम/न्यूनतम मानलाई local maxima/minima भनेर चिनिन्छ।

Please note that:

1. The combination of maxima and minima is extrema.
2. global maxima is he highest point on the curve within the given range and minima is the lowest point on the curve.
3. local maxima can be less than local minima and vice-versa

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How to compute Maxima/Minima

1. Step 1: Find the turning point:
   Evaluate f ′(x) and solve f ′(x)=0.
   Say p=a,b,c,... are solutions (these are turning points).
2. Step 2: Evaluate f"(x) and find f"(p).
   If f "(p) < 0 then at x= p maxima, and f(p) is maximum value
   If f "(p)> 0 then at x=p minima, and f(p) is minimum value
   If f "(p)= 0 then test fails at x=p
   
3. Repeat this for all critical points p=a,b,c,...

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Activity 1

1. Find the local minimum and maximum values of the function
   1. \(f(x) =x^3-6x^2-36x + 4\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x^3-6x^2-36x + 4\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 3x^2- 12x- 36\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 6x- 12\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 3x^2- 12x- 36 = 0\)
      or \( x^2- 4x- 12 = 0\)
      or \( x = 6, x= - 2\)

      Now, the analytical table is given below.

      	x =-2	x = 6
      \( f''(x) = 6x- 12\)	f''(-2)=-24 (-ve)	f''(6)=24 (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x^3-6x^2-36x + 4\)	f(-2)=44	f(6)=-212

      
      The function f(x) has maximum value at x=-2, the maximum value is f(-2)=44.
      The function f(x) has minimum value at x=6, the minimumm value is f(6)=-212.
      
   2. \(f(x) =3x^4-4x^3-12x^2+5\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = 3x^4-4x^3-12x^2+5\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 12x^3-12x^2-24x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 36x^2-24x-24\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 12x^3-12x^2-24x = 0\)
      or \( 12x(x^2-x-2) = 0\)
      or \( x = 0,x= - 1,x=2\)

      Now, the analytical table is given below.

      	x =-1	x = 0	x = 2
      \( f''(x) = 36x^2-24x-24\)	f''(-1)=36 (+ve)	f''(0)=-24 (-ve)	f''(2)=72 (+ve)
      Result	Minimum value	Maximum value	Minimum value
      \( f(x) = 3x^4-4x^3-12x^2+5\)	f(-1)=0	f(0)=5	f(2)=-27

      
      The function f(x) has minimum value at x=-1, the minimumm value is f(-1)=0
      The function f(x) has maximum value at x=0, the maximum value is f(0)=5
      The function f(x) has minimum value at x=2, the minimumm value is f(2)=-27
      
   3. \(f(x) =x +2 \sin x;0 \le x\le 2 \pi\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x +2 \sin x\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 1 +2 \cos x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = -2 \sin x\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 1 +2 \cos x = 0\)
      or \(\cos x = -\frac{1}{2}\)
      or \( x = \frac{2 \pi}{3}, x= \frac{4 \pi}{3}\)

      Now, the analytical table is given below.

      	\( x = \frac{2 \pi}{3}\)	\( x = \frac{4 \pi}{3}\)
      \( f''(x) = -2 \sin x\)	\(f''(\frac{2 \pi}{3})=-\sqrt{3}\) (-ve)	\(f''(\frac{4 \pi}{3})=\sqrt{3}\) (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x +2 \sin x\)	\(f(\frac{2 \pi}{3})=3.83\)	\(f(\frac{4 \pi}{3})=2.46\)

      
      The function f(x) has maximum value at \( x = \frac{2 \pi}{3}\), the maximum value is \(f(\frac{2 \pi}{3})=3.83\)
      The function f(x) has minimum value at \( x = \frac{4 \pi}{3}\), the minimumm value is \(f(\frac{4 \pi}{3})=2.46\)
      
   4. \(f(x) =x^3-5x+3\)
   5. \(f(x) =\frac{x}{x^2+4}\)

      Solution 👉 Click Here

      Given function is
      \( f(x) =\frac{x}{x^2+4}\)
      Differentiating the function w.r. to x, we get
      \( f'(x) =\frac{4-x^2}{(x^2+4)^2}\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( \frac{4-x^2}{(x^2+4)^2} = 0\)
      or \( x = -2, x= 2\)

      Now, the analytical table is given below.

      	\( x =-2\)	\( x =2\)
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)	\(f''(-2)=1/16\) (+ve)	\(f''(2)=-1/16\) (-ve)
      Result	Minimum value	Maximum value
      \( f(x) = \frac{x}{x^2+4}\)	\(f(-2)=-1/4\)	\(f(2)=1/4\)

      
      The function f(x) has maximum value at \( x = 2\), the maximum value is \(f(2)=1/4\)
      The function f(x) has minimum value at \( x =-2\), the minimumm value is \(f(-2)=-1.4\)
      
   6. \(f(x) =x+\sqrt{1-x}\)

   Solution 👉 Click Here

   Given function is
   \( f(x) =x+\sqrt{1-x}\)
   Differentiating the function w.r. to x, we get
   \( f'(x) =1-\frac{1}{2\sqrt{1-x}}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \( 1-\frac{1}{2\sqrt{1-x}} = 0\)
   or \( x =3/4\)
   Now, the analytical table is given below.
   	\( x =3/4\)
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)	\(f''(3/4)=-2\) (-ve)
   Result	Maximum value
   \( f(x) = x+\sqrt{1-x}\)	\(f(3/4)=1.25\)
   
   The function f(x) has maximum value at \( x = 3/4\), the maximum value is \(f(3/4)=1.25\)
   
2. Show that \(f(x)= x^3-3x + 4\) is maximum when \( x =-1\) , minimum when \( x = 1\) , neither when \( x = 0\) .

   Solution 👉 Click Here

   Given function is
   \( f(x) = x^3-3x + 4\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 3x^2-3\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =6x\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(3x^2-3 = 0\)
   or \( x^2 = 2\)
   or \( x = -1, x= 1\)

   Now, the analytical table is given below.

   	x =-1	x = 1
   \( f''(x) = 6x\)	f''(-1)=-6 (-ve)	f''(1)=6 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x^3-3x + 4\)	f(-1)=6	f(1)=2

   
   The function f(x) has maximum value at x=-1, the maximum value is f(-1)=6.
   The function f(x) has minimum value at x=1, the minimumm value is f(1)=2.
   
3. Show that the maximum value of a function \( f (x) = x + \frac{100}{x}-25\) is less than its minimum value.

   Solution 👉 Click Here

   Given function is
   \( f(x) = x + \frac{100}{x}-25\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 1-\frac{100}{x^2}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =\frac{200}{x^3}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(1-\frac{100}{x^2} = 0\)
   or \( x^2 = 100\)
   or \( x = -10, x= 10\)

   Now, the analytical table is given below.

   	x =-10	x = 10
   \( f''(x) = \frac{200}{x^3}\)	f''(-10)=-0.2 (-ve)	f''(10)=0.2 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x + \frac{100}{x}-25\)	f(-10)=-45	f(10)=-5

   
   The function f(x) has maximum value at x=-10, the maximum value is f(-10)=-45.
   The function f(x) has minimum value at x=10, the minimumm value is f(10)=-5.
   
4. Show that the maximum value of a function \( f (x) = x^{5/3}-5x^{2/3}\) at x=0 is 0. Also find its minumum value at x=2.

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Activity 2

Find the maximum and minimum values of following functions

1. \(f(x)=2x^3-3x^2-12x\)
2. \( f(x) = \frac{4}{3}x^3 + 5x^2- 6x-2\)
3. \(f(x)=2+3x-x^3\)
4. \(f(x)=200+8x^3+x^4\)
5. \(f(x)=x^4-6x^2\)
6. \(f(x)=3x^5-5x^3+3\)
7. \(f(x)=(x^2-1)^3\)
8. \(f(x)=x \sqrt{x^2+1}\)
9. \(f(x)=x-3x^{1/3}\)
10. \(f(x)=\frac{1+x^2}{1-x^2}\)
11. \(f(x)=\frac{x}{(x-1)^2}\)
12. \(f(x)=\sqrt{x^2+1}-x\)

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Activity 3

1. Find the two numbers whose sum is 16 and the sum of whose squares is minimum.

   Solution 👉 Click Here

   Let \( x\) and \( y\) be two numbers such that
   \( x + y = 16\)
   If \( S\) be the sum of the squares of the two numbers, then
   \( S = x^2 + y^2\)
   or \( S = x^2 + (16- x)^2\) \( (y = 16- x)\) भएकोले
   Differentiating the function S w.r. to x, we get
   \( \frac{dS}{dx} = 4x- 32.\)
   Again, differentiating the function S' w.r. to x, we get
   \( \frac{d^2S}{dx^2} = 4\)
   For stationary point, set
   \( \frac{dS}{dx} = 0 \)
   or \( 4x- 32 = 0\)
   or \( x = 8 \)
   Since \( \frac{d^2S}{dx^2 } = 4 > 0\) , so \( S\) is minimum when \( x = 8\)
   Now, the corresponding value of \( y\) is
   \( y = 16- 8 = 8\)
   Hence the required numbers are \( 8,8\)
2. Find the dimensions of a rectangle with perimeter 100 metres so that the area of the rectangle is a maximum.
3. A rectangular box with a square base and no top is to have a volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.
4. A farmer has 24 m of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
5. A farmer with 80m of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
6. A rectangular storage container With an open top is to have a volume of \(36 m^3\). The length of its base is twice the width. Material for the base costs Rs2 per square meter. Material for the sides costs Rs16 per square meter. Find the cost of materials for the cheapest such container.
7. Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
8. A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
9. A right circular cylinder is inscribed in a right circular cone so that the center lines of the cylinder and the cone coincide. The cone has 8 cm and radius 6 cm. Find the maximum volume possible for the inscribed cylinder.
10. Suppose a company’s daily profit \(f(q)=100q-0.1q^2\) is the quantity of calculators sold per day. How many calculators should the company sell to maximise profit and what is the maximum daily profit?