Question

discuss the maximum or minimum of the function:- 2xy-3x^2y-y^3+x^3y+xy^3​

Answer 1

Concept:

A function is a relation between a set of inputs having one output each.

A function from a set X to a set Y assigns to each element of X exactly one element of Y.

Given:

We are given the function:

2xy-3x²y-y³+x³y+xy³​

Find:

We need to find the minimum or maximum of the function.

Solution:

f(x, y)=2xy-3x²y-y³+x³y+xy³​

Differentiating with respect to x:

=2y-6xy+3x²y+y³  ...(1)

Differentiating with respect to y:

=2x-3x²-3y²+x³+3xy²     ...(2)

Now equating both (1) and (2)

2y-6xy+3x²y+y³ =2x-3x²-3y²+x³+3xy²

x³-y³+3xy²-3x²y-3x²-3y²+6xy+2x-2y=0

(x-y)³-3(x-y)²+2(x-y)=0

(x-y)((x-y)²-3(x-y)+2)=0

(x-y)(x²+y²-3x+3y-2xy+2)=0

So, if (x-y)>0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)<0 and (x²+y²-3x+3y-2xy+2)>0, then the function will be minimum.

And if (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0, then the function will be maximum.

Therefore, the function is minimum when (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0 and the function is maximum when (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0.

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Answer 2

Concept:

A function is a relation between a set of inputs having one output each.

A function from a set X to a set Y assigns to each element of X exactly one element of Y.

Given:

We are given the function:

2xy-3x²y-y³+x³y+xy³​

Find:

We need to find the minimum or maximum of the function.

Solution:

f(x, y)=2xy-3x²y-y³+x³y+xy³​

Differentiating with respect to x:

=2y-6xy+3x²y+y³  ...(1)

Differentiating with respect to y:

=2x-3x²-3y²+x³+3xy²     ...(2)

Now equating both (1) and (2)

2y-6xy+3x²y+y³ =2x-3x²-3y²+x³+3xy²

x³-y³+3xy²-3x²y-3x²-3y²+6xy+2x-2y=0

(x-y)³-3(x-y)²+2(x-y)=0

(x-y)((x-y)²-3(x-y)+2)=0

(x-y)(x²+y²-3x+3y-2xy+2)=0

So, if (x-y)>0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)<0 and (x²+y²-3x+3y-2xy+2)>0, then the function will be minimum.

And if (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0, then the function will be maximum.

Therefore, the function is minimum when (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0 and the function is maximum when (x-y)<0 and (x²+y²-3x+3y-2xy+2)<0 or (x-y)>0 and (x²+y²-3x+3y-2xy+2)>0.

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