To find the maximum and minimum turning points of y = f(x), we need to find x such that f'(x) = 0
Step 1 :
Draw a large, clear diagram for the situation.
Step 2 :
Construct the equation with the variable to be maximized or minimized as the subject of the formula in terms of the single variable x.
Step 3 :
Find the first derivative and equate to 0 to get the critical number.
Step 4 :
Apply the value that we have received as critical number in the second derivative.
Step 5 :
To find the maximum or minimum value, we have to apply the value of x in the original function.
Problem 1 :
When a small business employs x workers to manufacture its goods, the profit made is given by
P(x) = -2x^{3} + 2400 x - 4000 euros per week.
a) How many employees should they use to maximise profit ?
b) What is the maximum profit ?
Solution :
P(x) = -2x^{3} + 2400 x - 4000
P'(x) = -2(3x^{2}) + 2400(1) - 0
P'(x) = -6x^{2} + 2400
P'(x) = 0
-6x^{2} + 2400 = 0
-6x^{2} = - 2400
x^{2} = 400
x = 20
Finding the second derivative :
P'(x) = -6x^{2} + 2400
P''(x) = -6(2x) + 0
= -12x
When x = 20, P''(20) = -12(20)
= -240 < 0
Then maximum at x = 20
a) Number of employees required to maximize the profit is 20.
b) P(x) = -2x^{3} + 2400 x - 4000
P(20) = -2(20)^{3} + 2400 (20) - 4000
= -2(8000) + 48000 - 4000
= -16000 + 48000 - 4000
= 28000
The maximum profit is 28000 euros.
Problem 2 :
Square corners are cut from a piece of 12 cm by 12 cm tinplate which is then bent to form an open dish. What size squares should be removed to maximise the capacity of the dish ?
Solution :
Capacity of the dish = length x width x height
length = 12 - 2x, width = 12 - 2x and height = x
Capacity = (12 - 2x)(12 - 2x) x
f(x) = x(12 - 2x)^{2}
f'(x) = x(2(12-2x))(-2) + (1) (12 - 2x)^{2}
= -4x (12-2x) + (12 - 2x)^{2}
f'(x) = (12 - 2x) [-4x + 12 - 2x]
= (12 - 2x) [-6x + 12]
f'(x) = 0
(12 - 2x) (12 - 6x) = 0
x = 6, x = 2
f'(x) = (12 - 2x) [-6x + 12]
f'(x) = (12 - 2x) (-6) + (-6x + 12) (-2)
= -72 + 12x + 12x - 24
f''(x) = - 86 + 24x
Applying the values of x one by one.
When x = 2 f''(2) = - 86 + 24(2) = -86 + 48 = -38 < 0 maximum |
When x = 6 f''(6) = - 86 + 24(6) = -86 + 144 = 58 > 0 minimum |
The maximum capacity occurs when x = 2 cm, we should cut out 2 cm squares.
Problem 3 :
An open rectangular box has a square base, and its outer surface area must be 108 cm^{2}
a) Find the equation in terms of x and y.
b) Write y in terms of x.
c) Find the formula for the capacity C in terms of x and y.
d) What size must the base be in order to maximise the capacity ?
Solution :
a) Surface area of the square base cube = 4a^{2}
Area of four faces + area of bottom face = 108
4xy + x^{2} = 108
b) Solving for y in terms of x,
4xy = 108 - x^{2}
y = (108 - x^{2})/4x
y = (108/4x) - (x^{2}/4x)
y = (27/x) - (x/4)
c) Capacity of the cube = length x width x height
= x ⋅ x ⋅ y
= x^{2}[(27/x) - (x/4)]
= 27x - (x ⋅x^{2} /4)
= 27x - (x^{3}/4)
d) f(x) = 27x - (x^{3}/4)
f'(x) = 27 - 3x^{2}/4
Put f'(x) = 0
27 - 3x^{2}/4 = 0
3x^{2}/4 = 27
x^{2} = 27(4/3)
x^{2} = 36
x = 6
f''(x) = 0 - 6x/4
f''(x) = -3x/2
f''(6) = -18/2
= -9 < 0 maximum
At x = 6, the area will be maximised.
Problem 4 :
The slant edge of the cone has length 12 cm. If the cone has height x cm, find
a) an expression for the volume of the cone in terms of x.
b) The value of x for which the cone has the maximum possible volume.
Solution :
Here height of the cone is x cm.
Volume of cone = 1/3 π r^{2} h
a) Volume of cone = 1/3 π r^{2} x
r^{2} = 12^{2} - x^{2}
r^{2} = 144 - x^{2}
V (x) = 1/3 π(144 - x^{2}) ⋅ x
V (x) = 1/3 π(144x - x^{3})
b) V (x) = 1/3 π(144x - x^{3})
V' (x) = 1/3 π(144 - 3x^{2})
V' (x) = 0
1/3 π(144 - 3x^{2}) = 0
144 - 3x^{2 }= 0
3x^{2 }= 144
x^{2 }= 48
x = 4√3
Applying x = 4√3 in second derivative,
V'' (x) = 1/3 π(0 - 6x)
V'' (x) = -2xπ
V'' (4√3) = -2(4√3)π
= -8√3π < 0 maximum
To find the maximum value, we apply x = 4√3 in the original function.
V (x) = 1/3 π(144(4√3) - (4√3)^{3})
= 1/3 π(576√3 - 192√3)
= 1/3 π(384√3)
= 128π
Maximizing the volume at x = 4√3 and the volume is 128π cubic units.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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