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Find the extrema of f(x, y) = 4x + 2y, subject to the constraint 2x^2 + 3y^2 = 189.

maximum: ? at (x,y) = ( , )

minimum: ? at (x,y) = ( , )

2 Answers

2 votes

Answer:

.

Explanation:

User J Kluseczka
by
5.8k points
4 votes

Answer:

maximum at (x,y) = (1/9, 1/27)

minimum at (x,y) = (- 1/9, - 1/27)

Explanation:

The Method of Lagrange Multipliers is used when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant)

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

where:

f(x, y) = 4x + 2y

g(x,y)=2x^2 + 3y^2 = 189

Then:

(4,2)=λ(4x,6y)

2x^2 + 3y^2 = 189

---------------------

4=4λx ⇒ x=1/λ

2=6λy ⇒ y=1/3λ

2x^2 + 3y^2 = 189

---------------------

We replace x and y in the third equation:

2*(1/λ)^2+3*(1/3λ)^2=189 / *3λ^2

6+1=567λ^2

λ=
√(81) \\

λ=±9

---------------------

λ=+9

(x,y)=(1/9, 1/27) ⇒ f(x,y) >0 ⇒ (x,y) is a maximum

---------------------

λ= -9

(x,y)=(-1/9,- 1/27) ⇒ f(x,y) <0 ⇒ (x,y) is a minimum

User Colin Zhong
by
6.1k points