Applications of Derivatives Test

Result

Applications of Derivatives Test - 9

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Question 1
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A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the radius of the semicircular opening of the window to admit maximum light.

A

B

C

D

Solution
Question 2
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The point of local maxima for the function f(x) = sinx. cos x is

A
π/3

B
π/2

C
π/4

D
π/6

Solution

f(x) = sinx.cosx
f ’(x) = -sin² x + cos² x
-sin² x + cos² x = 0
sin² x = cos² x
tan² x = 1
tanx = +-1
for tan x = 1, x = π/4
for tan x = -1, x = 3 π/4
At x = π/4, f(π/4) = ½
At x = 3 π/4, f(3 π/4) = -½
At π/4 is the local maxima.
Question 3
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A real number x when added to its reciprocal give minimum value to the sum when x is

A
1/2

B
-1

C
1

D
2

Solution
Question 4
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The sum of two positive numbers is 20. Find the numbers if their product is maximum

A
5,15

B
8,12

C
20,0

D
10,10
Question 5
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The function f(x) = log x

A
Has a local maximum but no local minimum value

B
Has both ,a local minimum and a local maximum value

C
Has neither a local minimum nor a local maximum value

D
Has a local minimum but no local maximum value

Solution

f(x) = log x
f ’(x) = 1/x
Putting f ’(x) = 0
1/x = 0
x = ∞(this is not defined for x)
So, f(x) does not have minima or maxima
Question 6
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f(x) = x⁵ –5x⁴ + 5x³ –1. The local maxima of the function f(x) is at x =

A
1

B
5

C
0

D
3

Solution

f(x) = x⁵ - 5x⁴ + 5x³ - 1
On differentiating w.r.t.x, we get
f '(x) = 5x⁴ - 20x³ + 15x²
For maxima or minima,f '(x) = 0
⇒5x⁴ - 20x³ + 15x² = 0
⇒5x² (x² -4x+3) = 0
⇒5x² (x-1)(x-3) = 0
⇒x = 0,1,3
So, y has maxima at x = 1 and minima at x = 3
At x = 0, y has neither maxima nor minima, which is the point of inflection .
f(1) = 1-5+5-1=0,which is local maximum value at x = 1.
Question 7
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How many units should be sold so that a company can make maximum profit if the profit function for x units is given by p(x) = 25 + 64x - x²

A
16 units

B
32 units

C
8 units

D
64 units
Question 8
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Find two numbers whose sum is 24 and product is a large as possible.

A
12 and 12

B
20 and 4

C
8 and 16

D
None of the above
Question 9
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A point c in the domain of a function f is called a critical point of f if

A
f ’(x) = 0 at x = c

B
Either f ’(c) = 0 or f is not differentiable

C
f is not differentiable at x = c

D
f ”(x) = 0, at x = c

Solution

A point C in the domain of a function f at which either f(c) = 0 or f is not differentiable.
The point f is called the critical point.
c is called the point of local maxima
If f ′(x) changes sign from positive to negative as x increases through c, that is, if f ′(x) >0 at every point sufficiently close to and to the left of c, and f ′(x) <0 at every point sufficiently close to and to the right of c.
c is called the point of local minima
If f ′(x) changes sign from negative to positive as x increases through c, that is, if f ′(x) <0 at every point sufficiently close to and to the left of c, and f ′(x) >0 at every point sufficiently close to and to the right of c.
c is called the point of inflexion
If f ′(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.
Question 10
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Find two positive numbers x and y such that x + y = 60 and xy³ is maximum

A
x = 45, y = 15

B
x = 15, y = 45

C
x = 10, y = 50

D
x = 30, y = 30

Solution

two positive numbers x and y are such that x + y = 60.
x + y = 60
⇒x = 60 –y ...(1)
Let P = xy³
∴P =(60 –y)y³ = 60y³ –y4
Differentiating both sides with respect to y, we get

For maximum or minimum dP/dy = 0
⇒180y² - 4y³ = 0
⇒4y² (45 - y) = 0
⇒y = 0 or 45 - y = 0
⇒y = 0 or y = 45
⇒ y = 45 (∵y = 0 is not possible)

Thus, the two positive numbers are 15 and 45.