Hi all, please help me with this question.It's topic on limits and partial derivative. Find the maximal value of `f(x,y)=xy(1-x^2/a^2-y^2/b^2)^0.5` for `a=74.4` , `b=64.8` Round off your answer to...

Hi all, please help me with this question.It's topic on limits and partial derivative.

Find the maximal value of `f(x,y)=xy(1-x^2/a^2-y^2/b^2)^0.5` for `a=74.4` , `b=64.8` Round off your answer to 4 decimal places.

Asked on by authrun123

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tiburtius | High School Teacher | (Level 2) Educator

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First you need to find partial derivatives:

`(delf)/(delx)=y sqrt[1 - x^2/a^2 - y^2/b^2]-(x^2 y)/(a^2 sqrt[1 - x^2/a^2 - y^2/b^2])`

`(delf)/(dely)=x sqrt[1 - x^2/a^2 - y^2/b^2]-(x y^2)/(b^2 sqrt[1 - x^2/a^2 - y^2/b^2])`

Now you find stationary points (where partial derivatives are equal to 0). And afrer we rewrite our derivatives as product we get:

`sqrt(1-x^2/a^2-y^2/b^2)(y-(x^2y)/(a^2(1-x^2/a^2-y^2/b^2)))=0`

`sqrt(1-x^2/a^2-y^2/b^2)(x-(xy^2)/(b^2(1-x^2/a^2-y^2/b^2)))=0`

Obviously one solution would be `sqrt(1-x^2/a^2-y^2/b^2)=0`

hence `x^2/a^2+y^2/b^2=1`

But that's not the solution because if we look at our function

`f(x,y)=xy sqrt(1-x^2/a^2-y^2/b^2)`

we see that `x^2/a^2+y^2/b^2=1` is actually the edge of the functions domain (if we had `x^2/a^2+y^2/b^2>1` we would get square root of a negative number).

So we need to solve

`y-(x^2y)/(a^2(1-x^2/a^2-y^2/b^2))=0`

`x-(xy^2)/(b^2(1-x^2/a^2-y^2/b^2))=0`

Since a fraction is equal to 0 only if numerator is equal to 0 we have:

`y(a^2(1-x^2/a^2-y^2/b^2)-x^2)=0`

`x(b^2(1-x^2/a^2-y^2/b^2)-y^2)=0`

One solution is obviously `x_1=0,` `y_1=0.` ` `

Remaining four solutions we get by solving

`(a^2(1-x^2/a^2-y^2/b^2)-x^2)=0`

`(b^2(1-x^2/a^2-y^2/b^2)-y^2)=0`

Now we get

`a^2-x^2-(y^2a^2)/b^2-x^2=0`

`b^2-(x^2b^2)/a^2-y^2-y^2=0`

From first equation we have `x^2=(a^2(1-y^2/b^2))/2`

now we put that into second equation

`b^2-(a^2(1-y^2/b^2)b^2)/2a^2=2y^2`

`b^2-b^2/2+y^2/2=2y^2`

`y=pm b/sqrt(3)`

now we return that to our substitution to get

`x^2=(a^2(1-(b^2/3)/b^2))/2=(2/3 a^2)/2=a^2/3`

`x=pm a/sqrt3`

So stationary points are `T_1(0,0),\ T_2(-a/sqrt3,-b/sqrt3),\ T_3(-a/sqrt3,b/sqrt3),\ T_4(a/sqrt3,-b/sqrt3),`

` T_5(a/sqrt3,b/sqrt3)`

Now we need to find hessian `H=[[(del^2f)/(delx^2),(del^2f)/(delxdely)],[(del^2f)/(delydelx),(del^2f)/(dely^2)]]`

If `H` is positive definite then we have minimum and if `H` is negative definite we have maximum.

`(del^2f)/(delx^2)=`

`-x y (x^2/(a^4 (1 - x^2/a^2 - y^2/b^2)^(3/2)) + 1/( a^2 sqrt[1 - x^2/a^2 - y^2/b^2]))-((2 x y)/(a^2 sqrt[1 - x^2/a^2 - y^2/b^2]))`

`(del^2f)/(dely^2)=`

`-x y (y^2/(b^4 (1 - x^2/a^2 - y^2/b^2)^(3/2)) + 1/( b^2 sqrt[1 - x^2/a^2 - y^2/b^2]))-((2 x y)/(b^2 sqrt[1 - x^2/a^2 - y^2/b^2]))`

`(del^2f)/(delxdely)=(del^2f)/(delydelx)=`

`-((x^2 y^2)/(a^2 b^2 (1 - x^2/a^2 - y^2/b^2)^(3/2))) - x^2/( a^2 sqrt[1 - x^2/a^2 - y^2/b^2])` `- y^2/( b^2 sqrt[1 - x^2/a^2 - y^2/b^2]) + sqrt[1 - x^2/a^2 - y^2/b^2]`

Now you would put each of your stationary points to see the values of `H.`

I will skip that, because it's very long but not vrey difficult, and tell you the solution.

Minimums are points `T_3` and `T_4.`

`T_1` is sadle point.

Maximums are points `T_2(-42.9549,-37.4123)`

and `T_5(42.9549,37.4123)`

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