3. Brief about theory of maxima minima:
In mathematical analysis, the maxima and minima of a function, known collectively
as extrema are the largest and smallest value of the function, either within a
given range or on the entire domain
Pierre de Fermat was one of the first mathematicians to propose a general
technique, adequality, for finding the maxima and minima of functions.
As defined in set theory, the maximum and minimum of a set are the greatest and
least elements in the set, respectively. Unbounded infinite sets, such as the set of real
numbers, have no minimum or maximum.
4. The area of a rectangular lot, for example, could be determined based on the cost of
fencing and the property’s length and width. As a result, A = f(x) can express the area.
The most common issue is to figure out what value of x will result in the highest value
of A. We set dA/dx = 0 to find this value.
For 1st order and one variable:
Maximum-Let f(x) be a real function defined on an interval [a,b]. Then f(x) is said to
have the maximum value in [a,b], if there exist a point c in [a,b] such that f(x) ≤ f (c)
for all x ∈ [a,b]
Minimum- Let f(x) be a real function defined on an interval [a,b].Then, f(x) is said to
have the minimum value in [a,b], if there exist a point c in [a,b] such that f(x) ≥ f(c) for
all x ∈ [a,b]
5. Applications of maxima minima:
• There are numerous practical applications. Such applications exist in
economics, business, and engineering.
For example, in any manufacturing business it is usually possible to
express profit as a function of the number of units sold. Finding a
maximum for this function represents a straightforward way of
maximizing profits.
In other cases, the shape of a container may be determined by
minimizing the amount of material required to manufacture it. The
design of piping systems is often based on minimizing pressure drop
which in turn minimizes required pump sizes and reduces cost. The
shapes of steel beams are based on maximizing strength.
7. For 2nd order and functions of two variable:
z = f(x, y) ; x and y are the independent variables and z is the
dependent variable.
1.z=f(x,y)
2. find and del f/del y
3. del f/del x=0, del f/del y =0 and solv for x and y
4.
5. if rt-s^2>0 and r<0 then f has maximum value at that point.
6. if rt-s^2>0 and r>0 then f has minimum value at that point.
7. if rt-s^2<0 then f has neither maxima nor minima (saddle point)
8. if rt-s^2=0 then no conclusion.
𝜕𝑓
𝜕𝑥
= 0
8. EXAMPLE 1-
• Q.Find the maximum and minimum values of 𝒙𝟑
+ 𝟑𝒙𝒚𝟐
− 𝟑𝒙𝟐
− 𝟑𝒚𝟐
+ 𝟒.
• STEPS TO SOLVE-
• Find
𝜕𝑓
𝜕𝑥
,
𝜕𝑓
𝜕𝛾
,
𝜕2𝑓
𝜕𝑥2 ,
𝜕2𝑓
𝜕𝑦2 ,
𝜕2𝑓
𝜕𝑥𝜕𝑦
.
• Solve the equations
𝜕𝑓
𝜕𝑥
= 0 and
𝜕𝑓
𝜕𝛾
= 0 .
• Find all the values of x and y and the coordinates.
• Find r=
𝜕2𝑓
𝜕𝑥2 , s=
𝜕2𝑓
𝜕𝑥𝜕𝑦
, and t=
𝜕2𝑓
𝜕𝑦2.
• Check 𝑟𝑡 − 𝑠2
for every value of x and y.
• if 𝑟𝑡 − 𝑠2 > 0 and r<0 then f has maximum value at that point.
if 𝑟𝑡 − 𝑠2
> 0 and r>0 then f has minimum value at that point.
if 𝑟𝑡 − 𝑠2
< 0 then f has neither maxima nor minima (saddle point).
if 𝑟𝑡 − 𝑠2
= 0 then no conclusion.
• Substitute value of x and y to find the maximum value of 𝑓 𝑥 .
9. EXAMPLE 2-
Q. A garden is to be laid out in a rectangular area and protected by a chicken wire
fence. What is the largest possible area of the garden if only 100 running feet of
chicken wire is available for the fence ?
• STEPS TO SOLVE-
• Take variables x and y for length and breadth of the rectangle.
• Form two equations with area and parameter of the rectangle.
• Substitute value of x in terms of y from second equation.
• Express the third equation as area in terms of x.
• Obtain the fourth equation by differentiate the third equation w.r.t x.
• Equate it to zero as to find the value of x.
• Differentiate the fourth equation again w.r.t x to check the maxima.
• Substitute the value of x in equation 4 to get the largest possible area.
