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Examples : Graphical method Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Graphical method Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is the graphical method? |  |
Ans. The graphical method is a technique used to solve optimization problems in mathematics and economics. It involves graphing the constraints and objective function of a problem to visually determine the optimal solution.
| 2. How does the graphical method work? | |
Ans. The graphical method works by plotting the constraints of a problem on a graph and finding the feasible region, which is the area where all constraints are satisfied. Then, the objective function is plotted, and the optimal solution is found by identifying the point where the objective function is maximized or minimized within the feasible region.
| 3. What are the advantages of using the graphical method? | |
Ans. The graphical method has several advantages. Firstly, it provides a visual representation of the problem, making it easier to understand and analyze. Secondly, it is relatively simple and requires minimal calculations, making it suitable for problems with few variables. Lastly, it allows for quick identification of the optimal solution by visually inspecting the graph.
| 4. What are the limitations of the graphical method? | |
Ans. While the graphical method is useful for solving simple optimization problems, it has some limitations. It can only solve problems with a small number of variables, as graphing becomes complex and time-consuming with more variables. Additionally, it may not be accurate for problems with non-linear constraints or objective functions, as it relies on linear relationships.
| 5. Can the graphical method be used for real-world applications? | |
Ans. Yes, the graphical method can be applied to real-world problems. It is commonly used in fields such as economics, operations research, and engineering to analyze and optimize various scenarios. However, for more complex problems, computer-based optimization techniques are often preferred to handle larger datasets and non-linear relationships efficiently.
Text Transcript from Video
hello friends this video on linear
programming part 4 is bachi by exam
clear calm no more fear from exam before
watching this video please make sure
that you have lost part 1 part 3
let's take some more example here so
area to maximize sorry mini why is this
guy minimize this guy says that
development has strange so let's draw
this X plus T y is equal to 3 that is X
is equal to 0 Y is equal to 1 and Y is
equal to 0 X infinity so let's plot this
point 1 2 3 4 1 2 3 X is equal to 0 Y is
equal to 1 this point X is equal to 3 y
squared is 0 so I can draw one line
using the sponsor so this is nothing but
my X plus 3 y is equal to Z let's find
which area it will form let us put 0
comma 0 here so if you put 0 comma 0
this becomes 0 plus 0 is greater than 3
no that means this area is not part of
this line so I am looking for this area
correct this is the second is X plus y
is equal to 2 so X plus y is equal to 3
you put X equal to 0 y is equal to X is
equal Y is equal to 0 input you get X
equal to 2 so 0 comma 2 is 1 point and
this guy's so let me draw a line on this
this is my life so this is X plus y is
equal to B now let's put 0 comma 0 here
if you put 0 comma 0 here you get 0 plus
0 that is not greater than 2 that means
0 comma 0 is not part of this here also
I am looking for this reason this is
done x and y both of it isn't 0 that is
this guy and so if you now find a common
area the comedy I will get is this
because my X is greater than 0 and Y is
greater than 0 this is my con is not Y
because this this guy satisfy this area
this there satisfy this area X is
greater than Azalea Y is greater than
incidence this is my area I am looking
for if you see this is unbounded
unbounded so in this case solution may
exist or may not exist right but what
are the points the points here are this
guy is one point this guy is one point
and this guy's right so my point is
zero comma 2 is one point right this is
3 comma 0 is another point and one more
point I have this point is intersection
of these two line so let's find the
intersection of these two lines so I can
do that X plus 3y is equal to 3 and X
plus y is equal to subtract this you get
Y is 2 y is 1 or you get Y as invited if
Y is on by 2 my X will be 2 minus y that
is 2 minus 1 by 2 that is 3 right that
means my another point is 3 by 2 N 1 by
2 right so this point is 3 by 2 for ma 1
by this one so I have all these points
that is this point is 0 comma 2 this
mine this point is 3 comma 0 at this
point so let's find the corner point and
the value of Z so this is 0 2 3 0 3 by 2
or this minds think is 0 2 my Z is
nothing but 3x plus 5y right so this
becomes 3 into 0 plus 5 into 2 10 4 3
comma 0 this becomes 3 into 3 9 plus 5
into 0 there is 9 for this 3 by 2 in 1
by 2 this becomes 3 into 3 by 2 does
mine by 2 plus 5 into 1 by 2 there is my
plus 5 by 2 working by 2 it is said I am
looking for minimum so my value of M is
7 this may be Maxima
this may be minima or may not be now we
have to find find that I have sold that
I have to form this equation 3x plus 5y
is less than M M is
I have to find this equation and then I
have to say with that this equation has
any conflicting region with this so for
this 3x + 5 y is less than same and
let's find them
so here you put X is equal to zero you
get Y is 7 by 5 that is 124 you put Y is
0 you get x s 7 by 3 right that is
nothing but 2.3 oh yeah 2.3 so let's
plot the value X is 1.4 is somewhere
here X is 0 so x is 0 Y is 1.4 is
somewhere here and Y is 0 and X is 2.3
somewhere here right so if you draw this
line and you want to find Michigan this
belong less than 7 so let's take 0 + 0 0
plus 0 is 1 yes so I am looking for this
reason and if you see that this
particular reason is not contracting
with this yield there is no interference
no interference or you can see that no
common point since there is no
interference I can say that my 7 is
minimum correct you see what how our
done I want to find it 7 this minima or
not I drew this line and I saw there is
no common interference between this
region my the new region and the old
reason and then I can see that 7 is done
in the same example maximum would not
have worked because maximum would have
been 10 they would have be reformed the
equation 3x plus 5y is greater than 10
and this line you see if you actually
plot this line
we'll be in this region this direction
and if you see that there will be a
overlapping mccrane this reasons which
you get from this line and this line so
that you can say that the minima for
this function is there available but the
Maxima for this function is not
available and this is something which
you can tell by looking at the clock
only because if you see the graph is
something like this it is going in this
direction right so it can have any value
of x and y and for any positive value of
x and y this will always grow z will
always grow you see Z will keep doing
correct so you can see that just by
looking at the graph you can see that
the Maxima is not available for this
because it's keeps growing
but the memer may exist and we tried and
we saw that the minima actually exists
it for this condition let us take one
more example we have X plus 2y near
defined this is my objective function
and the conditions are 2x plus y is 8
and 3 X plus 2 y is greater than 6 and
X's and Y's confusing so let's form the
inflation of this guy this guy 2x plus y
is greater than 3 so for this way X plus
y is equal to 3 you put X is equal to 0
you get Y is equal to 3 you get Y is
equal to 0 you get excess 1.5 so let us
plot this point 1 2 3 4 1 2 so x is 0 Y
is 3 y is 0 X is 1 point 5 so we'll plot
this noise
this is my graph for 2x plus y is equal
right now I can find which period is
pointing to let's put 0 comma 0 here
this becomes 0 plus 0 is greater than 3
no so that means I am looking for this
region correct
second is X plus 2y is equal to 6 X plus
2 y is equal to 6 you put X equal to 0
you get Y is equal to 3 you put Y is
equal to 0 you get x into 6 so Y is
equal to 0 X is 4 5 6 somewhere here and
X is equal to 0 Y is 1 into here so we
have this pan right this guy's X plus 2y
which region this region this region
let's put 0 0 but 0 0 this becomes 0 is
greater than 6 no that means I'm looking
for this reason only right
so you find a common between this
distance in this reason also X is
greater than 0 that is this reasons why
is greater than 0 this reason then you
see that the reason I am looking for is
this guy's these two guys I'm looking
for all these values here are something
am and if you extend this further I am
looking for only for these values right
the the corner points here are at this
point just guy this guy all right so the
corner point is guy is actually if you
see is there is 6 comma 0 and this guy
is 0 comma 3
so my corner point is 0 comma 3 and 6 0
let's find the value of Z in this X plus
2y so this becomes 6 0 plus 23 and this
becomes 6
and the minimum value is if you see is
six only so you can take any of this now
I have to maximize this sorry minimize
this right so I have to this is M it may
be minimal not may not even Emma so I
have to first form it as equation X plus
2y is less than M and then check whether
this is overlapping and M is here 6 this
is over length so X plus 2y is less than
6 is nothing but then you can take the
same equation X plus 2y is greater than
6 less than 6 is this part and if you
see there is no overlap better in this
area and that area since there is no
overlap I can see that yes many my exist
and the minimum value is 6 and if you
see that this exists in all the point in
this line please note that minima occurs
at all points on line X plus y is 6
right you see that the minimize existed
at all of this mind if I take any line
in this any point in this line the value
is 6 itself so this is one example where
the minimize is not only on the end
points it exists in the end point also
like at these points also now let's say
matter here we'll take an example here
to maximize and minimize both this
function subject to these conditions
let's see this thank you visit exam
viacom to watch free a GP should be
doers try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and sorry
programming part 4 is bachi by exam
clear calm no more fear from exam before
watching this video please make sure
that you have lost part 1 part 3
let's take some more example here so
area to maximize sorry mini why is this
guy minimize this guy says that
development has strange so let's draw
this X plus T y is equal to 3 that is X
is equal to 0 Y is equal to 1 and Y is
equal to 0 X infinity so let's plot this
point 1 2 3 4 1 2 3 X is equal to 0 Y is
equal to 1 this point X is equal to 3 y
squared is 0 so I can draw one line
using the sponsor so this is nothing but
my X plus 3 y is equal to Z let's find
which area it will form let us put 0
comma 0 here so if you put 0 comma 0
this becomes 0 plus 0 is greater than 3
no that means this area is not part of
this line so I am looking for this area
correct this is the second is X plus y
is equal to 2 so X plus y is equal to 3
you put X equal to 0 y is equal to X is
equal Y is equal to 0 input you get X
equal to 2 so 0 comma 2 is 1 point and
this guy's so let me draw a line on this
this is my life so this is X plus y is
equal to B now let's put 0 comma 0 here
if you put 0 comma 0 here you get 0 plus
0 that is not greater than 2 that means
0 comma 0 is not part of this here also
I am looking for this reason this is
done x and y both of it isn't 0 that is
this guy and so if you now find a common
area the comedy I will get is this
because my X is greater than 0 and Y is
greater than 0 this is my con is not Y
because this this guy satisfy this area
this there satisfy this area X is
greater than Azalea Y is greater than
incidence this is my area I am looking
for if you see this is unbounded
unbounded so in this case solution may
exist or may not exist right but what
are the points the points here are this
guy is one point this guy is one point
and this guy's right so my point is
zero comma 2 is one point right this is
3 comma 0 is another point and one more
point I have this point is intersection
of these two line so let's find the
intersection of these two lines so I can
do that X plus 3y is equal to 3 and X
plus y is equal to subtract this you get
Y is 2 y is 1 or you get Y as invited if
Y is on by 2 my X will be 2 minus y that
is 2 minus 1 by 2 that is 3 right that
means my another point is 3 by 2 N 1 by
2 right so this point is 3 by 2 for ma 1
by this one so I have all these points
that is this point is 0 comma 2 this
mine this point is 3 comma 0 at this
point so let's find the corner point and
the value of Z so this is 0 2 3 0 3 by 2
or this minds think is 0 2 my Z is
nothing but 3x plus 5y right so this
becomes 3 into 0 plus 5 into 2 10 4 3
comma 0 this becomes 3 into 3 9 plus 5
into 0 there is 9 for this 3 by 2 in 1
by 2 this becomes 3 into 3 by 2 does
mine by 2 plus 5 into 1 by 2 there is my
plus 5 by 2 working by 2 it is said I am
looking for minimum so my value of M is
7 this may be Maxima
this may be minima or may not be now we
have to find find that I have sold that
I have to form this equation 3x plus 5y
is less than M M is
I have to find this equation and then I
have to say with that this equation has
any conflicting region with this so for
this 3x + 5 y is less than same and
let's find them
so here you put X is equal to zero you
get Y is 7 by 5 that is 124 you put Y is
0 you get x s 7 by 3 right that is
nothing but 2.3 oh yeah 2.3 so let's
plot the value X is 1.4 is somewhere
here X is 0 so x is 0 Y is 1.4 is
somewhere here and Y is 0 and X is 2.3
somewhere here right so if you draw this
line and you want to find Michigan this
belong less than 7 so let's take 0 + 0 0
plus 0 is 1 yes so I am looking for this
reason and if you see that this
particular reason is not contracting
with this yield there is no interference
no interference or you can see that no
common point since there is no
interference I can say that my 7 is
minimum correct you see what how our
done I want to find it 7 this minima or
not I drew this line and I saw there is
no common interference between this
region my the new region and the old
reason and then I can see that 7 is done
in the same example maximum would not
have worked because maximum would have
been 10 they would have be reformed the
equation 3x plus 5y is greater than 10
and this line you see if you actually
plot this line
we'll be in this region this direction
and if you see that there will be a
overlapping mccrane this reasons which
you get from this line and this line so
that you can say that the minima for
this function is there available but the
Maxima for this function is not
available and this is something which
you can tell by looking at the clock
only because if you see the graph is
something like this it is going in this
direction right so it can have any value
of x and y and for any positive value of
x and y this will always grow z will
always grow you see Z will keep doing
correct so you can see that just by
looking at the graph you can see that
the Maxima is not available for this
because it's keeps growing
but the memer may exist and we tried and
we saw that the minima actually exists
it for this condition let us take one
more example we have X plus 2y near
defined this is my objective function
and the conditions are 2x plus y is 8
and 3 X plus 2 y is greater than 6 and
X's and Y's confusing so let's form the
inflation of this guy this guy 2x plus y
is greater than 3 so for this way X plus
y is equal to 3 you put X is equal to 0
you get Y is equal to 3 you get Y is
equal to 0 you get excess 1.5 so let us
plot this point 1 2 3 4 1 2 so x is 0 Y
is 3 y is 0 X is 1 point 5 so we'll plot
this noise
this is my graph for 2x plus y is equal
right now I can find which period is
pointing to let's put 0 comma 0 here
this becomes 0 plus 0 is greater than 3
no so that means I am looking for this
region correct
second is X plus 2y is equal to 6 X plus
2 y is equal to 6 you put X equal to 0
you get Y is equal to 3 you put Y is
equal to 0 you get x into 6 so Y is
equal to 0 X is 4 5 6 somewhere here and
X is equal to 0 Y is 1 into here so we
have this pan right this guy's X plus 2y
which region this region this region
let's put 0 0 but 0 0 this becomes 0 is
greater than 6 no that means I'm looking
for this reason only right
so you find a common between this
distance in this reason also X is
greater than 0 that is this reasons why
is greater than 0 this reason then you
see that the reason I am looking for is
this guy's these two guys I'm looking
for all these values here are something
am and if you extend this further I am
looking for only for these values right
the the corner points here are at this
point just guy this guy all right so the
corner point is guy is actually if you
see is there is 6 comma 0 and this guy
is 0 comma 3
so my corner point is 0 comma 3 and 6 0
let's find the value of Z in this X plus
2y so this becomes 6 0 plus 23 and this
becomes 6
and the minimum value is if you see is
six only so you can take any of this now
I have to maximize this sorry minimize
this right so I have to this is M it may
be minimal not may not even Emma so I
have to first form it as equation X plus
2y is less than M and then check whether
this is overlapping and M is here 6 this
is over length so X plus 2y is less than
6 is nothing but then you can take the
same equation X plus 2y is greater than
6 less than 6 is this part and if you
see there is no overlap better in this
area and that area since there is no
overlap I can see that yes many my exist
and the minimum value is 6 and if you
see that this exists in all the point in
this line please note that minima occurs
at all points on line X plus y is 6
right you see that the minimize existed
at all of this mind if I take any line
in this any point in this line the value
is 6 itself so this is one example where
the minimize is not only on the end
points it exists in the end point also
like at these points also now let's say
matter here we'll take an example here
to maximize and minimize both this
function subject to these conditions
let's see this thank you visit exam
viacom to watch free a GP should be
doers try free online tests get the best
quality study materials study from the
best tutors and mentors and much more
thanks and sorry
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About this Video
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Nov 22, 2024 Last updated |
Video Description: Examples : Graphical method for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
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