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Applications of Differential Calculus
Mathematics
University of Hertfordshire
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Foundation Mathematics 2: Week 4
Applications of differential calculus
Spring 2021
Contents
- 1 Maximising and Minimising Functions
- 1 Stationary Points
- tive test 1 Locating stationary points, and classifying them usingthe second deriva-
- 1.2 Example
- 1 First derivative test
- 1.3 Example
- 1 Stationary Points
- 2 Maximisation and Minimisation Problems
- 2 Motivation
- 2 General principles
- 2 Worked examples
- 2.3 Example
- 2.3 Example
- 2.3 Example
- 3 Summary
1 Stationary Points
Iff′(x) = 0 whenx=afor some valuea, thenais called a statonary point or turning point of the functionf. There are three main types:
(a) Local maximum. e. the point (0,0) ofy=−x 2. (b) Local minimum. e. the point (0,0) ofy=x 2.
(c) Point of inflection. e. the point (0,0) ofy=x 3.
We use the terms “local” max/min, to distinguish from the global max/min which is the overall extreme value for the given range.
Example of local and global extrema:
1.2 Example
Find and classify the stationary points of
y=x 3 +x 2
Solution:
The first derivative is:
y′= 3x 2 + 2x
Setting this equal to zero and solving forxto locate the stationary points:
y′= 0
3 x 2 + 2x= 0
x(3x+ 2) = 0
∴x= 0 or 3x+ 2 = 0
∴x= 0 or x=−
2
3
Then determining they-coordinate in each case:
y(x= 0) = 0 3 + 0 2 = 0
and
y
(
x=−
2
3
)
=
(
−
2
3
) 3
+
(
−
2
3
) 2
= −
8
27
+
4
9
=
12 − 8
27
=
4
27
So the stationary points are (0,0) and
(
− 23 , 274
)
.
Then we differentiate again to obtain the second derivative:
y′′= 6x+ 2
andevaluatingthisatthestationarypointstoapplythesecondderivativetestineachcase:
y′′(x= 0) = 6(0) + 2 = 2> 0
So (0,0) is a local minimum.
y′′
(
x=−
2
3
)
= 6
(
−
2
3
)
+ 2 =−4 + 2 =− 2 < 0
So
(
− 23 , 274
)
is a local maximum.
1.3 Example
Find and classify the stationary points of
f(x) =x 4 + 2x 3
Solution:
The first derivative is:
f′(x) = 4x 3 + 6x 2
Setting this equal to zero and solving forxto locate the stationary points:
f′(x) = 0
4 x 3 + 6x 2 = 0
2 x 3 + 3x 2 = 0
x 2 (2x+ 3) = 0
∴x 2 = 0 or 2x+ 3 = 0
∴x= 0 or x=−
3
2
Then determining they-coordinate in each case:
f(x= 0) = 0 4 + 2(0) 3 = 0
and
f
(
x=−
3
2
)
=
(
−
3
2
) 4
+ 2
(
−
3
2
) 3
=
81
16
−
27
4
= −
27
16
So the stationary points are (0,0) and
(
− 32 ,− 2716
)
.
Then we differentiate again to obtain the second derivative:
f′′(x) = 12x 2 + 12x
andevaluatingthisatthestationarypointstoapplythesecondderivativetestineachcase:
f′′(x= 0) = 12(0) 2 + 12(0) = 0
So the second derivative test fails in this case.
f′′
(
x=−
3
2
)
= 12
(
−
3
2
) 2
+ 12
(
−
3
2
)
=
12 × 9
4
−
36
2
= 27− 18
= 9> 0
So
(
− 32 ,− 2716
)
is a local minimum.
Now, we can apply the first derivative test to classify (0,0) by evaluatingf′(x) around x= 0:
- Tryx=− 0 .0001:
f′(x=− 0 .0001) = 4(− 0 .0001) 3 + 6(− 0 .0001) 2
= 5. 9996 × 10 − 8 > 0
- Tryx= +0:
f′(x=− 0 .0001) = 4(0) 3 + 6(0) 2
= 6. 0004 × 10 − 8 > 0
Hence we have a positive point of inflection at (0,0).
2 Maximisation and Minimisation Problems
2 Motivation
Next, we shall learn how to use interpret written physical problems, and use the techniques of differential calculus to obtain solutions that maximise or minimise a quantity.
We have learned that differentiation can be used to find maximum or minimum values of functions.
Motivation: maximising profits in a company, or the surface area of a solar panel, or minimising the amount of material used in production of an item.
We will be looking for the largest or smallest value of a function subject to some kind of constraint.
The constraint will be some condition (usually described byan equation) that must absolutely be true no matter what our solution is.
For example, we might need to minimise the material used in making an oil drum, but it must have a particular capacity no matter what dimensions we choose.
2 General principles
Begin by reading the question, twice!
We need to clearly understand what are (a) the quantity to be optimised, and (b) the constraint that must be satisfied.
It often helps to draw a diagram of the situation.
Assign some variable names to the unknowns, and turn our constraint and the optimised quantity into equations.
We will want to use the constraint to obtain a formula in one variable for the quantity to be optimised.
Then we can find the choice for which it is maximised or minimised by differentiating this formula and setting the derivative equal to zero.
Finally,usethefirstorsecondderivativetesttoconfirmthattheanswerisspecifically a maximum or minimum as desired.
2.3 Example 2
A farmer wants to erect a rectangular pen in his field. He has 40mof wire fencing. What dimensions should he use to make the pen as large as possible?
- Ask yourself: What is the quantity to be optimised? What is theconstraint?
The optimisation is that we need tomaximisethe area of the pen. The constraint is that theperimetermust equal the fencing length of 40m. 2. Draw a rectangle to visualise the problem.
- Name the unknowns: the lengthxand breadthyof the pen.
- Ask yourself: Using these symbols, how can the constraint and the optimised quan- tity be expressed as equations?
- Use the information provided to create a formula for the constraint: the perimeter P= 40 of the pen.
P=x+y+x+y= 2x+ 2y= 40 =⇒ x+y= 20
- Write a formula for what we are trying to optimise: the areaAof the pen.
A=xy
- Use the constraint equation to eliminate a variable from the area formula:
y= 20−x =⇒ A=x
(
20 −x
)
= 20x−x 2
- Now that the area is stated in terms of one variable, differentiate: dA dx
= 20− 2 x
Set the gradient equal to zero and solve to find the value ofxthat maximises the area A: dA dx = 0 =⇒ 20 − 2 x= 0 =⇒ 2 x= 20 =⇒ x= 10m
Use the constraint to determine the other variable, so that the full dimensions are known:
y= 20−x= 20−10 = 10m
- Confirm that this choice (x= 10m, y = 10m) maximises the product using the Second Derivative Test: d 2 A dx 2
=−2 =⇒
d 2 A dx 2
∣
∣∣
∣
x=
=− 2 < 0
Hence this is a local maximum area.
- What is the largest pen he can make? Substitute both valuesback into the formula for area to determine what this maximum area actually is:
x= 10, y= 10 =⇒ A=xy= 10×10 = 100m 2
- Then substitute in the volume restriction to eliminateh:
S= 2πr 2 + 2000r− 1
Differentiate this function: dS dr = 4πr− 2000 r− 2
SetddSr = 0 and solve forrto determine the stationary point of the surface area function:
4 πr− 2000 r− 2 = 0 =⇒ 4 πr 3 −2000 = 0
Hence:
∴r 3 =
2000
4 π =⇒ r= 3
√
2000
4 π = 5. 419 cm
- Use the value ofrto calculatehand thus obtain the full dimensions of the drum:
h=
1000
π(5) 2 = 10. 84 cm
- Use the Second Derivative Test to confirm that this is a minimum:
d 2 S dr 2
= 4π+ 4000r− 3 =⇒ d 2 S dr 2
∣∣
∣
∣r=5. 419 = 4π+ 4000(5)
− 3 = 37. 70 > 0
Hence this is a local minimum surface area.
3 Summary
- Given a functiony= f(x), the local maximum or minimum value ofyoccurs at stationary points, where: dy dx
= 0
- Given thatx=ais a stationary point ofy=f(x), we can determine whether it is a maximum or minimum using the Second Derivative Test:
d 2 y dx 2
∣∣
∣
∣x=a
>0 =⇒ Local Minimum. <0 =⇒ Local Maximum. = 0 =⇒ Use the First Derivative Test.
Applications of Differential Calculus
Module: Mathematics
University: University of Hertfordshire
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