find the extremum of f(x,y) subject to the given constraint, and state wether it is a maximum or minimum.
4y^2-29x^2; 3x+y=21
Review your text section (probably Lagrange multipliers), and see whether you can come up with this solution:
http://www.wolframalpha.com/input/?i=minimize+4y%5E2-29x%5E2+subject+to+3x%2By%3D21
To find the extremum of the function \(f(x,y) = 4y^2 - 29x^2\) subject to the constraint \(3x + y = 21\), we can use the method of Lagrange multipliers.
Step 1: Solve the constraint equation for one variable in terms of the other. In this case, solve for \(y\):
\(3x + y = 21 \Rightarrow y = 21 - 3x\)
Step 2: Substitute this expression for \(y\) into the objective function \(f(x,y)\) to express it solely in terms of \(x\):
\(f(x,y) = 4(21 - 3x)^2 - 29x^2\)
Step 3: Compute the partial derivatives of \(f\) with respect to \(x\) and \(y\):
\(\frac{{\partial f}}{{\partial x}} = -24(21 - 3x) - 58x\)
\(\frac{{\partial f}}{{\partial y}} = 8(21 - 3x)\)
Step 4: Set up the system of equations by equating the partial derivatives to zero, introducing a Lagrange multiplier \(λ\):
\(\frac{{\partial f}}{{\partial x}} = λ\frac{{\partial g}}{{\partial x}}\)
\(\frac{{\partial f}}{{\partial y}} = λ\frac{{\partial g}}{{\partial y}}\)
where the constraint function \(g(x,y)\) is \(g(x,y) = 3x + y - 21\).
Substituting the partial derivatives and the constraint equation into the system, we have:
\(-24(21 - 3x) - 58x = λ(3)\)
\(8(21 - 3x) = λ(1)\)
Step 5: Solve the system of equations simultaneously to find the values of \(x\), \(y\), and \(λ\). From the first equation, we get:
\(-72x + 504 - 58x = 3λ \Rightarrow -130x = 3λ - 504\)
From the second equation, we get:
\(168 - 24x = λ\)
Setting these two equations equal to each other, we can find \(x\):
\(-130x = 3(168 - 24x) - 504\)
\(-130x = 504 - 72x - 504\)
\(-58x = 0\)
\(x = 0\)
Now, substitute this value of \(x\) back into the second equation to find \(λ\):
\(168 - 24(0) = λ\)
\(168 = λ\)
Step 6: Using the values of \(x\), \(y\), and \(λ\), find the corresponding values for the objective function and classify the extremum.
When \(x = 0\), substituting into the constraint equation: \(3(0) + y = 21\)
\(y = 21\)
So the extremum occurs when \(x = 0\) and \(y = 21\), and we can substitute these values into the objective function to find the extremum value:
\(f(0,21) = 4(21)^2 - 29(0)^2\)
\(f(0,21) = 4(441) - 0\)
\(f(0,21) = 1764\)
Therefore, the extremum of \(f(x,y)\) subject to the constraint \(3x+y=21\) is a maximum, and its value is 1764.