Maxima & Minima MCQ Quiz in मल्याळम - Objective Question with Answer for Maxima & Minima - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

A function f(x) is called increasing on an interval I if given any two numbers, x_1, and x₂ in I such that x₁ < x2, we have f(x₁) < f(x₂).

Similarly, function f(x) is called decreasing on an interval I if given any two numbers, x1, and x2 in I such that x1 < x2, we have f(x1) > f(x2).

Calculation:

Here f(x) = min (4x + 3, x + 4) is an increasing function for every x ϵ [0, 3]

It means the value of f(x) increases as x increases.

∴ At x = 3, f(x) will be maximum

⇒ f(x) _max = min (4 × 3 + 3, 3 + 4)

⇒ f(x) max = min (15, 7)

⇒ f(x) max = 7

Hence the maximum value of the function is 7.

Explanation:

f(x, y) = x³y² (1 - x - y)

\(\frac{{\partial f}}{{\partial x}} = 3{x^2}{y^2} - 4{x^3}{y^2} - 3{x^2}{y^3}\)

\(\frac{{\partial f}}{{\partial y}} = 2{x^3}y - 2{x^4}y - 3{x^3}{y^2}\)

\(\frac{{{\partial ^2}f}}{{\partial {x^2}}} = r = 6x{y^2} - 12{x^2}{y^2} - 6x{y^3}\)

\(\frac{{{\partial ^2}f}}{{\partial x \cdot \partial y}} = s = 6{x^2}y - 8{x^3}y - 9{x^2}{y^2}\)

\(\frac{{{\partial ^2}f}}{{\partial {y^2}}} = t = 2{x^3} - 2{x^4} - 6{x^3}y\)

Now,

Put \(\frac{{\partial f}}{{\partial x}} = 0,\) we get x²y² (3 - 4x - 3y) = 0

\(\frac{{\partial f}}{{\partial y}} = 0,\)

We get x³y (2 - 2x - 3y) = 0

Solving these two, we get stationary points \( \to \left( {\frac{1}{2},\frac{1}{3}} \right)\;\& \;\left( {0,\;0} \right)\) 

Now,

rt - s² = x⁴y² [12 (1 - 2x - y)(1 - x - 3y) - (6 - 8x - 9y)²]

Since x, y ϵ (0, ∞)

Thus, for \(\left( {\frac{1}{2},\frac{1}{3}} \right)\) 

\(rt - {s^2} = \frac{1}{{16}}\frac{1}{9}\left[ {12\left( { - \frac{1}{3}} \right)\left( { - \frac{1}{2}} \right) - {{\left( {6 - 4 - 3} \right)}^2}} \right]\) 

\(rt - {s^2} = \frac{1}{{14}} > 0\)

\(\& \;r = 6\left( {\frac{1}{2} \cdot \frac{1}{9} - \frac{2}{4} \cdot \frac{1}{9} - \frac{1}{2} \cdot \frac{1}{{27}}} \right) = - \frac{1}{9} < 0\)

Hence, f(x, y) has a maxima at \(\left( {\frac{1}{2},\frac{1}{3}} \right)\) 

\(\therefore {\rm{Maximum\;value}} = \frac{1}{8} \cdot \frac{1}{9}\left( {1 - \frac{1}{2} - \frac{1}{3}} \right) = \frac{1}{{432}}\)

h² = a² + b² [Pythagoras theorem for right-angled triangle]

Given a + h = constant

Let it be K, i.e.

a + h = K

Area of triangle is given by:

\(A = \frac{1}{2} \times a \times b\) 

\(A = \frac{1}{2} \times a \times \sqrt {{h^2} - {a^2}} \) 

\(A = \frac{1}{2}a\sqrt {{{\left( {K - a} \right)}^2} - {a^2}} \) 

\(A = \frac{1}{2}a\sqrt {{K^2} - 2aK} \) 

For the maximum area of the triangle, we need to find the value of a and h where it will be maximum.

Taking the derivative of the above, and equating it to zero, we get:

\(\frac{{dA}}{{da}} = 0\) 

\( = \frac{1}{2}\sqrt {{K^2} - 2aK} + \frac{1}{2}a \times \frac{1}{2}\frac{{\left( { - 2K} \right)}}{{\sqrt {{K^2} - 2aK} }} = 0\) 

K² – 2aK – aK = 0

K² = 3 aK

K = 3a

a + h = K

a + h = 3a

h = 2a

\(\cos \theta = \frac{a}{h}\)

\(\cos \theta = \frac{a}{{2a}}\)

\(\cos \theta = \frac{1}{2}\)

θ = 60°

Explanation:

\(f\left( x \right) = 10{\left( {x - 1} \right)^{\frac23}}\)

Now, given that f(x) is a real-valued function.

If (x – 1) is negative, then the factional power of the negative number is imaginary.

So, (x – 1) cannot be negative.

Therefore, (x – 1) should be ≥ 0 and x ≥ 1.

f(x) is an increasing function. So, f(x) will be minimum at the minimum value of x.

Hence, f(x) will be minimum at x = 1.

Concept:

The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.

Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.

This is explained with the help of the following graph:

If \(\frac{{{d^2}f}}{{d{x^2}}} < 0\); the point will be a point of maxima

If \(\frac{{{d^2}f}}{{d{x^2}}} > 0\), the point will be a point of minima.

Calculation:

f(x) = ln (x + 1) – x

\(f'\left( x \right) = \frac{1}{{x + 1}} - 1\)

Equation f’(x) = 0, and solving for x, we get a possible maximum or minimum point, i.e.

\(\frac{1}{{x + 1}} - 1 = 0\)

x + 1 = 1

x = 0

Now, checking for f’’(x)|_{x = 0}, we get:

\(f''\left( x \right) = \frac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}\)

At x = 0,

\(f''\left( 0 \right) = \frac{{ - 1}}{{{{\left( {0 + 1} \right)}^2}}} = - 1\)

Since, f’’(0) < 0, x = 0 is a point of maximum.

∴ The maximum value of the function f(x) occurs at x = 0

Let the side of each of the squares cutoff be x m so that the height of the box is x m and the sides of the base are 6 – 2x, 2 – 2x m as shown in the figure.

Volume V of the box,

V = x (6 – 2x) (2 – 2x) = 4 (x3 – 4x2 + 3x)

\(\frac{{dv}}{{dx}} = 4\left( {3{x^2} - 8x + 3} \right)\)

For V to be maximum or minimum, we must have

\(\begin{array}{l} \frac{{dv}}{{dx}} = 0\\ \Rightarrow 3{x^2} - 8x + 3 = 0\\ \Rightarrow x = \frac{{8 \pm \sqrt {64 - 4\left( 3 \right)\left( 3 \right)} }}{6} = \frac{{8 \pm \sqrt {28} }}{6} = \frac{{4 \pm \sqrt 7 }}{3} \end{array}\)

⇒ x = 2.2 (or) 0.45 m

The value x = 2.2 m is not valid as no box is possible for this value. \(\frac{{{d^2}V}}{{d{x^2}}} = 4\left( {6x - 8} \right)\)

At x = 0.45 m, \(\frac{{{d^2}v}}{{d{x^2}}}\) is negative.

Hence at x = 0.45 m = 45 cms

Volume of the box is maximum.

\(f\left( x \right) = 2{x^2} - 2x + 6\)

\(f'\left( x \right) = 4x - 2\)

To find point at which maximum and minimum may occur equate:

\(f'\left( x \right) = 0\)

\(4x - 2 = 0\)

\(\therefore x = \frac{1}{2}\)

Since

\(f''\left( x \right) = 4 > 0\)

∴ at x = ½ minima can occur but not maxima

Also check the border condition for local maxima or minima

Therefore, maximum value is 10.

Concept:

Maxima and Minima of a function:

Let f(x, y) is supposed to be continuous for all values of x and y in the neighborhood of x = a and y = b. Then f(a,b) is called as maximum or minimum of f(x, y) according as f(a + h, b + k) is less than or greater than for all sufficiently small independent value of h and k, positive or negative, provided both of them are not equal to zero.

Necessary conditions for a point to be maxima or minima: The necessary conditions for f(x, y) to have maxima or minima at x = a, y = b is that

\({\frac{{\partial {\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial {\rm{x}}}}_{\left( {\begin{array}{*{20}{c}} {x = a,}\\ {y = b} \end{array}} \right)}} = 0{\rm{\;and\;\;}}{\frac{{\partial {\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial {\rm{y}}}}_{\left( {\begin{array}{*{20}{c}} {x = a,\;}\\ {y = b} \end{array}} \right)}} = 0\)

Sufficient conditions for a point to be maxima and minima:

\({\rm{r}} = {\frac{{{\partial ^2}{\rm{f}}}}{{\partial {{\rm{x}}^2}}}_{\left( {\begin{array}{*{20}{c}} {x = a,}\\ {y = b} \end{array}} \right)}};{\rm{\;s}} = {\frac{{{\partial ^2}{\rm{f}}}}{{\partial {\rm{x\;}}\partial {\rm{y}}}}_{\left( {\begin{array}{*{20}{c}} {x = a,}\\ {y = b} \end{array}} \right)}};{\rm{t}} = {\frac{{{\partial ^2}{\rm{f}}}}{{\partial {{\rm{y}}^2}}}_{\left( {\begin{array}{*{20}{c}} {x = a,}\\ {y = b} \end{array}} \right)}}{\rm{\;}}\)

Case 1. Maxima or Minima at x = a and y = b if rt > s².

if r > 0 → minima and if r < 0 → maxima

Case 2. Neither Maxima or Minima at x = a and y = b if rt < s².

The point is called saddle point.

Case 3. If rt = s² then further investigation needs to be done to know about the point

Calculation:

K(x, y) = 4x² + 6y² - 8x - 4y + 8

\(\frac{{\partial {\rm{K}}}}{{\partial {\rm{x}}}} = 8{\rm{x}} + 0 - 8 - 0 + 0 = 0\)

8x - 8 = 0 → x = 1

\(\frac{{\partial {\rm{K}}}}{{\partial {\rm{y}}}} = 0 + 12{\rm{y}} - 0 - 4 + 0 = 0\)

12y - 4 = 0 → y = 0.333

To check for maxima or minima,

\({\rm{r}} = \frac{{{\partial ^2}{\rm{K}}}}{{\partial {{\rm{x}}^2}}} = 8\)

\({\rm{s}} = \frac{{{\partial ^2}{\rm{K}}}}{{\partial {\rm{x\;}}\partial {\rm{y}}}} = 0\)

\({\rm{t}} = \frac{{{\partial ^2}{\rm{K}}}}{{\partial {{\rm{y}}^2}}} = 12\)

∵ r > 0 and rt > s²

∴ The function has minimum at x = 1 and y = 0.333

The minimum value of the function is:

\({\bf{K}}\left( {1,\;0.333} \right) = \frac{{10}}{3}\)

Calculation:

Given

\(f\left( {x,y,z} \right) = x + y + z\)

\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\)     --- (i)

\(\phi \left( {x,y,z} \right) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\)

Lagrange’s function

\(F\left( {x,y,z} \right) = f\left( {x,y,z} \right) + \lambda \times \phi \left( {x,y,z} \right)\)

∴ \(F\left( {x,y,z} \right) = x + y + z + \lambda \times \left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)\; \)

Lagrange’s equations are

\(\frac{{\partial F}}{{\partial x}} = 0\;,\frac{{\partial F}}{{\partial y}} = 0,\frac{{\partial F}}{{\partial z}} = 0\;\)

\(\frac{{\partial F}}{{\partial x}} = 1 + \lambda \times \left( { - \frac{1}{{{x^2}}}} \right)\; = 0\)

\(\lambda = {x^2}\) -- (ii)

\(\frac{{\partial F}}{{\partial y}} = 1 + \lambda \times \left( { - \frac{1}{{{y^2}}}} \right) = 0\;\;\)

\(\lambda = {y^2}\)     — (iii)

\(\frac{{\partial F}}{{\partial z}} = 1 + \lambda \times \left( { - \frac{1}{{{z^2}}}} \right) = 0\;\)

\(\lambda = {z^2}\)      --- (iv)

From (ii), (iii), (iv)

\(\lambda = {x^2} = {y^2} = {z^2}\)

\(i.e\;x = y = z = \surd \lambda \)

Substituting in equation (i), we get

\(\frac{1}{{\sqrt \lambda }} + \frac{1}{{\sqrt \lambda }} + \frac{1}{{\sqrt \lambda }} = 1\)

λ = 9

∴ \(x = y = z = \pm 3 \) 

∴ Stationary points are (3, 3, 3) and (-3,-3,-3)

But Stationary Point (-3,-3,-3) does not satisfy the constraint \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1\)

\(f\left( {x,y,z} \right) = x + y + z \)

∴ \(maximum\;value\;of\;f\left( {x,y,z} \right) = f\left( {3,3,3} \right) = 3 + 3 + 3 = 9 \)

Concept:

The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.

Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.

This is explained with the help of the following graph:

If \(\frac{{{d^2}f}}{{d{x^2}}} < 0\); the point will be a point of maxima

If \(\frac{{{d^2}f}}{{d{x^2}}} > 0\), the point will be a point of minima.

Calculation:

The given function \(f\left( x \right) = {x^4}{e^{ - \frac{{2x}}{3}}}\)

\(f'\left( x \right) = 4{x^3}{e^{ - \frac{{2x}}{3}}} - \frac{2}{3}{x^4}{e^{ - \frac{{2x}}{3}}}\)

To get the critical points, f’(x) = 0

\( \Rightarrow 4{x^3}{e^{ - \frac{{2x}}{3}}} - \frac{2}{3}{x^4}{e^{ - \frac{{2x}}{3}}} = 0\)

⇒ x = 0, 6

\(f''\left( x \right) = 4\left( {3{x^2}{e^{ - \frac{{2x}}{3}}} - \frac{2}{3}{x^3}{e^{ - \frac{{2x}}{3}}}} \right) - \frac{2}{3}\left( {4{x^3}{e^{ - \frac{{2x}}{3}}} - \frac{2}{3}{x^4}{e^{ - \frac{{2x}}{3}}}} \right)\)

At x = 0, f’’(0) = 0

At x = 6, f’’(6) < 0

Therefore, the given function has a maximum at a value of x = 6.