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October 23, 2014

Posted by **Carly** on Wednesday, August 10, 2011 at 3:05pm.

I solved fx=50x+20y+10

fy=6y+20x+4

gx=10x+4y

gy=4x

Then what?

- Calculus. Please help! -
**MathMate**, Wednesday, August 10, 2011 at 4:59pmI will solve it by Lagrange multipliers, although there may be other possible ways, depending on what you've done in school.

Let

f(x,y)=25x^2+3y^2+20xy+10x+4y+10 (function to be minimized)

c(x,y)=5x^2+4xy-1 (constraint = 0)

we will set

p(x,y)=f(x,y)+λ*c(x,y)

where λ is the Lagrange multiplier (lambda), to be determined.

We will calculate the partial derivatives px, py and equate to zero.

Together with the constraint c(x,y)=0, we have three equations in three unknowns (x,y,λ).

Solving the 3 non-linear equations will give x=1/sqrt(5), y=0, and λ=-sqrt(5)-5.

The minimum value of f(x,y) is 15+2√5.

Here are the details:

p(x,y)=(4xy+5x^2-1)λ+3y^2+20xy+4y+25x^2+10x+10

px(x,y)=(4y+10x)λ+20y+50x+10 = 0

py(x,y)=4xλ+6y+20x+4 = 0

c(x,y)=4xy+5x^2-1 = 0

Solve for x, y & λ from the three non-linear equations above to get

x=1/sqrt(5), y=0, λ=-sqrt(5)-5.

- Calculus. Please help! -
**Carly**, Wednesday, August 10, 2011 at 5:18pmWould I then plug in x and y into the original equation to find the minimum value? When I do that I get approximately 19.472, but my webwork homework says that is incorrect.

- Calculus. Please help! -
**MathMate**, Wednesday, August 10, 2011 at 6:02pmYou're right, the answer set

x=1/sqrt(5), y=0, λ=-sqrt(5)-5.

does not give the minimum.

I have taken the wrong set of answers which should have read:

x=-1/sqrt(5), y=0, λ=sqrt(5)-5

which gives f(x,y)=10.528 approx.

but unfortunately this one does not seem to be right either.

I'll look at it and get back to you.

In the mean time, you could repost and see if other teachers could give you an answer before I get back.

- Calculus. Please help! -
**MathMate**, Wednesday, August 10, 2011 at 6:16pmx=-1/sqrt(5), y=0, λ=sqrt(5)-5

which gives

f(x,y)=15-2√5

=10.528 approx.

should be the correct answer.

When I varied the values of x and y, I was able to get values of f(x,y) smaller than 15-2√5, which should not happen.

However, by varying the values of x and y, we no longer satisfy the constraints.

x=-1/sqrt(5), y=0 gives the minimum value when the constraints are satisfied.

- Calculus. Please help! -
**Anonymous**, Wednesday, August 10, 2011 at 8:46pmThat worked! Thank you very much. I spent a long time trying to solve that equation before I put it on jiskha. It was driving me crazy. I appreciated all your help!

- Calculus. Please help! :) -
**MathMate**, Wednesday, August 10, 2011 at 11:08pmYou're very welcome!

Please*do*rework the problem and understand every step. It may help your exams!

If there is any doubt, post again!

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