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MTH301 Calculus II Assignment No 01 Solution & Discussion Due Date: 27-05-2014

Total marks: 20
Lecture # 12 to 17 
Due date: May 27, 2014
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Q1: Find three positive numbers whose sum is 54 and whose product is as large as possible. Marks = 10

Q2: Let and R is the triangular region with vertices (0, 0), (2, 0) and (2, 2). Find the interior and boundary points only at which the absolute extrema of f(x, y) can occur. Marks = 10

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Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

plz mth 301 ki assiment ka solution upload kr de 

Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

dear i get (o,o) critical point in interior of the domain

it is possible that interior point could be 0,0

result how to get the value of the funtion f(0,0)

plz check answer is right
Question 1
ans f(x,y)=xy(54-x-y)
diff partially w.r.t 'x' and 'y' we get
f_x (x,y)=54y-2xy-((y)^(2)...­­­ (i)
f_y (x,y)=54-x-x^2-2xy...... (ii)
putting f_x=0,f_y=0
we get
x=18,y=18
since given that
x+y+z=54
so z=18
thus
x=18,y=18,z=18
Again partially diff w.r.t x,y
f_xx = -2y,f_xx(18,18)= -36
f_yy = -2x,f_yy(18,18)= -36
f_xy = 54-2x-2y f_xy(18,18)= -18
product will be max
D = f_xx (18,18) f_yy (18,18) f^2 (18,18)
D= (-36)(-36)-(18)^2
D= 1296-324
D= 972>0

plz mth 301 kiassiment 1 ka solution upload kre 

ab koi question 2 ka idea day den plzzzzzzzzzzzzz

kindly koi 2nd ques ka ans bta dy i have done but answer is not conform....

koi to solution bata de 2 question ka plzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

suppose vertices are       O(0,0),A(2,0),B(2,2)

For interior points with respect to x:

f(x)=(3x^2)-3y                                                     (a)

Interior points with respect y:

f(y)3-3x                                                                (b)

for the boundary points we take on the segment OA

y=0

U(x)=f(x,0)=x^3

Regarded as function of x denied on the closed interval (2<=x<=2) its extreme value may be occur at the end point x=2 and so on which corresponded to the point (2,0) and (2,2)

U’(x)=3x^2=0

On segment OB x=0 and V(y)=(0,y)=3y

Using symmetric form possible points are also these (0,0)(2,0)(2,2)

The interior point of AB y=2-x

Fy=x^3+3(2-x)-3x(2-x)

Fy=(x^3)-3(x^2)-9x+6

Take derivative

Fy’=3x^2-7x

 

 

Points are (2,2)

u did it wrong Aisha 


x^2+3y-3xy

fx(x,y)=2x-3y

fy(x,y)=3-3x

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