Maxima & Minima Concepts & Solved Examples for CAT - Nitin Gupta, AlphaNumeric


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    Concept : If the sum of positive numbers are constant then their product is maximum when all numbers are equal.

    Suppose that x, y, z . . . w are n positive variables and that c is a constant, then If x + y + z + ... + w = c, the value of xyz ... w is greatest when x = y = z = ... = w = c/n.

    a + b + c = 12 , find maximum value of a.b.c

    Ans : 64

    a + b + c =19, find maximum value of a.b.c

    Ans 19^3/27

    a + b + c = 7 ; find Maximum value of a^2 * b^3 * c^4

    Method 1:
    a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 = 7
    now a/2 = a/2 = b/3 = b/3 = b/3 = c/4 = c/4 = c/4 = c/4 = 7/9
    ===> a/2 × a/2 × b/3× b/3 × b/3 × c/4 × c/4 × c/4 × c/4 = (7/9)^9
    ===> a^2. b^3. c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

    Method 2 :
    AM ≥ GM
    a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 = 7
    {a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 } /9 = { a/2 × a/2 × b/3× b/3 × b/3 × c/4 × c/4 × c/4 × c/4 }^1/9
    Now on simplification we will get a^2 * b^3 * c^4 ≤ 2^2 × 3^3 ×4^4 × (7/9)^9

    Method 3( preferable)
    Here Powers of a, b,c are 2,3,4 respectively
    Divide 7 in the ratio 2:3:4
    ===> a =2/9 × 7; b= 3/9 × 7; c = 4/9 × 7
    Now on substitution you will get a^2 * b^3 * c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

    Method 4:
    To find the greatest value a^m * b^n * c^p . . . when a + b + c + ... is constant; m, n, p, ... being positive integers.
    Hence a^m *b^n * c^p ... will be greatest when the factors a/m , b/n , c/p are all equal, that is, when
    a/m = b/n = c/p = ... = (a+b+c+…….)/(m +n +p+….)
    here a/2 = b/3 =c/4 = 7/9
    now substitute the value & u will get
    a^2 * b^3 * c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

    If a, b and c are positive variables and a + b + c = 12, find maximum value of (a + 1) × (b + 2) × c.

    Here we need to find maximum value of (a + 1) × (b + 2) × c ; so we must know (a + 1) + (b + 2) + c
    Now (a + 1) + (b + 2) + c = 15 ; so each term (a + 1) = (b + 2) = c = 5
    So (a + 1) × (b + 2) × c = 125

    If a + 2b + 3c = 9 find maximum value of a × b × c. (Given a, b and c are positive).

    a = 2b =3c = 9/3 =3 ; so a * 2b * 3c = 27 ==> abc = 27/6

    If a + 2b + 3c = 63, find maximum value of a^3 × b^5 × c. (Given a, b and c are positive).

    a/3 = 2b /5 = 3c /1 = (a+2b+3c)/(3+5+1) = 63/9 = 7
    ==> a = 21; b= 35/2 ; c= 7/3
    So a^3 × b^5 × c = 21^3 * (35/2)^ 5 * 7/3

    Find the greatest value of (a + x)^3 * (a - x)^ 4 for any real value of x numerically less than a.

    Here a + x = p; a- x =q ;
    ==> p + q = 2a & we r supposed to find p^3 * q^4
    So p/3 = q/4 = 2a/7
    ==> p = 6a/7 & q = 8a /7

    If 2a + 5b = 7; find the maximum value of (a+1)^2 * (b+2)^3

    2a + 5b = 2(a+1) + 5 (b+2) = 2a+5b+12 = 19
    2x + 5y = 19 & we need to find x^2 * y ^3
    2x/2 = 5y/3 = 19/5 --
    x = 19/5 & y = 57 /25
    x^2 * y ^3 = (19/5)^2 * (57/25)^3
    = 19^5 * 3^3/5^8

    Concept: If the product of positive numbers are constant then sum is minimum when all numbers are equal.
    If xyz...w = c, the value of x + y + ... w is least when x = y =... = w , so that the least value of x + y + ... + w is nc^1/n.

    If a, b, c are positive and a × b × c = 125, find minimum value of a + b + c ?

    Ans : 15

    If a, b, c are positive and a × b × c = 17, find minimum value of a + b + c.

    Ans: 3*(17)^1/3

    Find the least value of 3x + 4y + 5z for positive values of x and y, subject to the condition xyz = 6.

    To find the minimum value of 3x + 4y + 5z ; we must know 3x × 4y × 5z
    So 3x × 4y × 5z = 3 * 4 * 5 * xyz = 360
    So 3x + 4y + 5z = 3 × (360)^1/3 = 6×(45)^1/3

    Find the least value of x + y + z for positive values of x, y & z subject to the condition x^2 * y^3 * z^4 = 6

    Method 1:
    x + y + z = x/2 + x/2 + y/3 +y/3 +y/3+ z/4+z/4+z/4+z/4 = ?
    so we must know the product of
    x/2 × x/2 × y/3 × y/3 ×y/3× z/4 × z/4 × z/4 × z/4 = (x^2×y^3×z^4)/ (2^2×3^3×4^4) = 6/(2^2×3^3×4^4)
    so x +y + z = 9 ×{ 6/(2^2×3^3×4^4)}^1/9

    Method 2:
    AM ≥ GM
    (x/2 + x/2 + y/3 +y/3 +y/3+ z/4+z/4+z/4+z/4)/9 ≥ (x/2 × x/2 × y/3 × y/3 ×y/3× z/4 × z/4 × z/4 × z/4)^1/9
    So , x + y + z ≥ 9 ×{ 6/(2^2×3^3×4^4)}^1/9

    Method 3
    If x +y +z =k & x^2 * y^3 * z^4 = 6
    So x/2 =y/3=z/4 = k/9
    So x = 2k/9; y =3k/9; z = 4k/9
    Now substitute the value of x,y & z in x^2 * y^3 * z^4 = 6
    K = 9 ×{ 6/(2^2×3^3×4^4)}^1/9

    Find the least value of 4x + 2y + 3z for positive values of x , y & z subject to the condition x^2 * y^3 * z^4 = 6.

    If 4x +2y +3z =k & x^2 * y^3 * z^4 = 6
    So 4x/2 =2y/3=3z/4 = k/9
    So x = k/18; y =k/6; z = 4k/27
    Now substitute the value of x,y & z in x^2 * y^3 * z^4 = 6

    If a * b * c = 100 , where a,b,c are positive integers , find the maximum & minimum value of a + b + c.

    We know that “if product is constant ,then sum is minimum when all no. are equal”
    But here if u take all equal then it won’t be an integer.
    So for the question based on integer --- if they cannot be equal then they must be very very close to each other , so 100 = 4 * 5 * 5
    So minimum value of x + y + z = 4 + 5 + 5 = 14;
    Now for maximum they must be far away from each other; so 100 = 1 * 1 * 100
    So max of x + y + z = 1 + 1 + 100 = 102

    If a + b + c = 100, where a, b and c are positive integers then find maximum & minimum value of a * b * c

    We know that “if sum is constant ,then product is maximum when all no. are equal”
    But here if u take all equal then it won’t be an integer.
    So for the question based on integer --- if they cannot be equal then they must be very very close to each other , so 100 = 34 + 33 + 33
    So maximum value of xyz = 34 * 33 * 33;
    Now for minimum they must be far away from each other; so 100 = 1+1+98
    So min of xyz = 1 * 1 * 98 =98

    Maxima & Minima in the context of Quadratic Equations

    Before going to maxima/minima questions, we will brush up some theory related to Quadratic equation

    Definition

    Any equation of degree 2 is known as a quadratic equation.
    General form is ax^2 + bx + c = 0
    The numbers a, b are called the coefficients of this equation and c is the constant.

    Roots

    The possible values of x which satisfy the quadratic equation are called the roots of the quadratic equation. Quadratic equation will have two roots either real or imaginary. We normally denote them as α and β.
    A root of the quadratic equation is a number such that pα ^2 + qα + r = 0 or pβ^ 2 + qβ + r = 0.

    If α & β are roots of the quadratic equation ax^2+bx + c = 0, then (x - α) and (x -β) are factors of ax^2+bx + c = 0
    OR ax^2 + bx + c = (x - α)(x -β)

    Properties of roots

    f(x) = ax^2 + bx + c = 0
    The sum of the roots= α + β = -b/a.
    The product of roots = α β = c/a
    If α and β are the roots of the equation ax^2 + bx + c = 0, then we can write
    ax^2 + bx + c = x^2 - (α + β) x + αβ = x^2 + (sum of the roots) x + product of the roots = 0
    Or, ax^2 + bx + c = a (x – α) (x – β) = 0

    Discriminant

    Given is the quadratic equation ax^2 + bx + c = 0, where a ≠ 0.
    (b^2 – 4ac) is also known as Discriminant (D)
    If D = 0, then √(b^2-4ac)= 0. So, the roots will be real and equal.
    If D > 0, then √(b^2-4ac)> 0. So, the roots will be real and distinct.
    If D < 0, then √(b^2-4ac) is not real. So, the roots will not be real.
    If D is a perfect square (including D = 0) and a, b and c are rational, then the roots will also be rational.

    Descartes' Rule of Signs of Roots

    The maximum number of positive roots of any equation is equal to the change of signs from positive (+ve) to negative (-ve) and from negative (-ve) to positive (+ve).

    Solving by factorization

    Ax^2 + Bx + C =0
    We have to write B as the sum of 2 numbers say P and Q such that the product of P and Q is equal to the product of A and C
    B = (P + Q)
    A * C = P * Q

    For example
    5x^2 -2x -4 = 0
    So P * Q = 5 * (-4) = -20
    And P + Q = -2
    By trial and error we get - 10 and 2
    5x^2 -10x + 2x -4 = 0
    5x(x-2) + 2(x-2) =0
    OR (5x+2)(x-2)=0

    Solving by Formula

    Assuming that α and β are the roots of the equation ax^2 + bx + c = 0, where a ≠ 0.
    Then α = (-b+√(b^2-4ac))/2a and β = (-b-√(b^2-4ac))/2a

    Minimum Value of Quadratic Equations

    Ax^2 + Bx + C = 0
    OR (sqrtA * x)^2 + (2 * (sqrtA * x) * B/(2 * (sqrtA)) ) + (B/(2 * (sqrtA * x)^2 + C = 0
    OR (sqrtA * x + (B/(2 * (sqrtA)) ^2 + (C - (B/(2 * (sqrtA * x)^2 ) = 0
    Now (sqrtA * x + (B/(2 * (sqrtA*x)) ^2 will always be positive as it’s a square term so its minimum value will be 0
    SO the minimum value of the expression (sqrtA * x + (B/(2 * (sqrtA * x)) ^2 + (C - (B/(2 * (sqrtA * x)^2 ) is equal to:
    (C - (B/(2 * (sqrtA * x)^2 )
    And this value is achieved when (sqrtA * x + (B/(2 * (sqrtA)) is 0
    Or x = -B/2A
    So the minimum value of a quadratic equation is (C - (B/(2 * (sqrtA * x)^2 ))
    And this value is achieved when x = -B/2A

    Now we will solve questions.

    Would the expression 18x – 3x^2 + 8 have a maxima or a minima?
    For what value of x would the maxima or minima occur?
    What is the maxima or minima value?

    Since a, the coefficient of x2 is negative, the expression would have a maxima value.
    The maxima value would occur when x = -b/2a = - 18/2 * -3 = 3
    To find the maxima value, substituting x = 3 in the expression we get 18 × 3– 3 × 3^2 + 8 = 54 – 27 + 8 = 35.

    Let p and q be the roots of the quadratic equation x^2 − (α − 2)x − α − 1 = 0. What is the minimum possible value of p^2 + q^2?

    (a - 2)^2 + 2a + 2
    = a^2 + 4 - 4a + 2a + 2
    = a^2 - 2a + 6
    = (a - 1)^2 + 5

    A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

    Maximum or Minimum Value of a Quadratic Function
    Let f be a quadratic function with standard form f (x) = a( x − ℎ )^2 + k.
    The maximum or minimum value of f occurs at x = ℎ.
    If a < 0, then the maximum value of f is f (ℎ) = k.
    So f (x) = a( x − 1 )^2 + 3.
    Now f(0) = 1
    So a = -2
    So the expression is -2(x-1)^2 + 3
    at x=10 -> - 2 * 9^2 + 3 = -159

    Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per piece up to 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

    let x batches of extra 20 samosas be there so
    size of box is (200+20x)
    price per box is (2-0.1x)
    revenue = (200+20x)(2-0.1x) (we need to maximize this)
    Now do not simplify it
    We know that maximum value exists at x = -b/2a = sum of roots/2
    So here sum of roots = -10 + 20 =10
    So revenue will be maximum when x = 10/2 = 5
    So maximum size of box is 200 + 10 * 5 = 300

    Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx^2, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
    How many units should Mr. David produce daily?
    What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

    1600c + 40b + 240 = 5/3 ( 400c + 20b + 240)
    4800c + 120b + 720 = 2000c + 100b + 1200
    2800c + 20b = 480
    so 140c + b = 24 -(1)
    and 3600c + 60b + 240 = 3/2 (1600c + 40b + 240)
    7200c + 120b + 480 = 4800c + 120b + 720
    2400c = 240 so c=1/10 so b=24 - 140 * 1/10 = 10

    so func is x^2/10 + 10x + 240
    profit = 30x - x^2/10 - 10x + 240 = -x^2/10 + 20x + 240
    so max at x = 100
    so 100 for 1st part.
    profit= -1000 + 2000 + 240 = 760

    Find the maximum and/or minimum value of (x^2 - x + 1)/(x^2 + x + 1) for real values of x.

    Standard way to solve this is
    Let (x^2-x+1)/(x^2+x+1) = k
    ⇒ x^2 − x +1= kx^2 + kx + k
    ⇒ (1− k ) x^2 − (1+ k ) x + (1− k ) = 0
    Now for x to be real, the determinant has to be greater than or equal to zero…
    (1+ k )^2 − 4 (1− k)( 1− k) ≥ 0
    ⇒ (1+ 2k + k^2 ) − 4(1− 2k + k^2 ) ≥ 0
    ⇒ −3k^ 2 +10k − 3 ≥ 0
    ⇒ 3k2 −10k + 3 ≤ 0
    ⇒ 3 ×(k − 3) × ( k -1/3) ≤ 0
    ⇒ 1/3 ≤ k ≤ 3
    So max value is 3 & minimum value is 1/3

    Find the maximum value of 1/(x^2 − 2x + 4)

    Same procedure as Q6. Answer would be 1/3

    How many integral values from 1 to 15 can the expression (x^2 + 34x -71)/(x^2 + 2x - 7) not take?

    (x^2 + 34x - 71)/(x^2 + 2x - 7) = k
    (1-k)x^2 + (34 - 2k)x - 71 + 7k = 0
    b^2 - 4ac > = 0 for real values therefore
    (34-2k)^2 - 4 * (1 - k)(-71 + 7k) > = 0
    k^2 - 14k + 45 > = 0
    that is K < = 5 and K > = 9 therefore k cannot take 3 values i.e 6,7,8

    If x + y + z = 5 ; xy + yz +zx = 3 then find the maximum & minimum value of x where x, y and z are real numbers.

    z = 5 - (x + y)
    xy + yz + zx = xy + 5(x + y) - (x + y)^2 = 3
    x^2 + x(y - 5) + (y^2 - 5y + 3) = 0
    D ≥ 0
    3y^2 - 10y - 13 ≤ 0
    -1 ≤ y ≤ 13/3

    else x + y + z = 5
    ie x= 5 - (y + z)
    also -> x(y + z) + yz = 3
    x(5 - x) + yz =3
    yz < = (y+z)^2/4 ( Am > Gm)
    ie yz < = (5-x)^2 /4
    so -> x(5-x) + (5-x)^2/4 < = 3
    (5-x)(4x+5-x) < = 12
    (5-x)(5+3x) < = 12
    -3x^2 + 10x + 25 < = 12
    ie -3x^2 + 10x + 13 < = 0
    (-1 , 13/3)

    The cost of fuel for running the engine of an army tank is proportional to the square of the speed and Rs. 64/ hour for a speed of 16 kmph. Other costs amount to Rs. 400/hour. the tank has to make a journey of 400 km at a constant speed.[JMET 2006]
    The most economial speed for this Journey is
    Find the cost at this most economical speed

    Cost of fuel is proportional to square of the speed.
    E = KS^2
    ⇒ 64 = K(16)^2
    ∴ K = 1/4
    Total cost = 1/4S^2* t + 400t
    But t = 400/s

    So total cost = 1/4S^2 * 400/s + 400 * 400/s
    == 100s + 160000/s
    Now at Most economical speed, the cost is minimum
    We know that sum of two no. is minimum if it product constant
    Here product = 100s * 160000/s =16000000
    So each term must be equal to 4000
    So 100s = 4000
    ⇒ S = 40
    ⇒ Cost = 4000+4000 = 8000

    If x and y are real numbers, then the minimum value of x^2 + 4xy + 6y^2 – 4y + 4 is [XAT 2010]

    x^2 + 4xy + 6y^2 – 4y + 4
    = x^2 + 4xy + 4y^2 + 2y^2 – 4y + 4
    = (x + 2y)^2 + 2(y^2 – 2y + 1 + 1)
    = (x + 2y)^2 + 2(y – 1)^2 + 2
    This expression will take the minimum value when both the square terms are zero
    ∴ (y – 1)^2 = 0
    ∴ y = 1
    and
    (x + 2y)^2 = 0
    ∴ x = –2
    ∴ The minimum value of this expression is 2

    Practice Questions

    1. A despatch rider is in open country at a distance of 6 kms from the nearest point P of a straight road..He wishes to proceed as quickly as possible to a point Q on the road 20kms from P..if his maximum speed across the country is 40 km/hr,then at what distance from P,he should strike the road?

    2. John has x children by his first wife..Mary has x + 1 children by her first husband. They marry and have children of their own. The whole family has 24 children. Assuming the two children of same parents do not fight,the find the maximum possible number of fights that can take place ?

    3. The circle x^2 + y^2 = 1 cuts the x axis at P and Q. Another circle with centre at Q and variable radius intersects the first circle at R above the x axis and line segment PQ at S. Find the maximum area of the triangle QSR?


 

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