Exercises for Lagrange Multipliers

In exercises 1-15, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

1) Objective function: \(f(x, y) = 4xy\) Constraint: \(\frac{x^2}{9} + \frac{y^2}{16} = 1\)

Answer:

Subject to the given constraint, the function \(f\) has a relative minimum of \(-24\) at both \( \left(-\frac{3\sqrt{2}}{2}, 2\sqrt{2}\right) \) and \( \left(\frac{3\sqrt{2}}{2}, -2\sqrt{2}\right) \) and a relative maximum of \(24\) at both \( \left(\frac{3\sqrt{2}}{2}, 2\sqrt{2}\right) \) and \( \left(-\frac{3\sqrt{2}}{2}, -2\sqrt{2}\right) \)

2) Objective function: \(f(x, y) = x^2y\) Constraint: \(x^2 + 2y^2 = 6\)

3) Objective function: \(f(x,y)=x^2 +y^2 +2x−2y+1\) Constraint: \( g(x,y)= x^2 +y^2 =2 \)

Answer:

Subject to the given constraint, \(f\) has a relative minimum of \(-1\) at \( (-1, 1) \) and a relative maximum of \(7\) at \( (1,-1) \).

4) Objective function: \(f(x, y) = xy\) Constraint: \(4x^2 + 8y^2 = 16\)

5) Objective function: \(f(x, y) = x^2 + y^2\) Constraint: \(xy = 1\)

Answer:

\(f\) has a relative minimum of \(2\) at both \( (-1, -1) \) and \( (1,1) \), subject to the given constraint.

6) Objective function: \(f(x, y) = x^2 - y^2\) Constraint: \(x−2y+6=0\)

7) Objective function: \(f(x, y) = x^2 + y^2\) Constraint: \(x+2y−5=0\)

Answer:

Subject to the given constraint,\(f\) has a relative minimum of \( f(1,2)=5\) at the point \( (1, 2) \).

8) Objective function: \(f(x, y) = x^2 + y^2\) Constraint: \((x−1)^2+4y^2=4\)

9) Objective function: \(f(x, y) = 4x^3 + y^2\) Constraint: \(2x^2 + y^2 = 1\)

Answer:

Subject to the given constraint, the function \(f\) has a relative minimum of \(-\sqrt{2}\) at \( \left(-\frac{\sqrt{2}}{2}, 0\right) \),
a relative minimum of \(\frac{25}{27}\) at both points \( \left(\frac{1}{3}, -\frac{\sqrt{7}}{3}\right) \) and \( \left(\frac{1}{3}, \frac{\sqrt{7}}{3}\right) \),
a relative maximum of \(\sqrt{2}\) at \( \left(\frac{\sqrt{2}}{2}, 0\right) \), and a relative maximum of \(1\) at both points \( (0,1) \) and \( (0,-1) \).

Solution:

Let \(g(x,y) = 2x^2 + y^2\) be the constraint function. Then:

\(\vecs\nabla f(x,y) = 12x^2 \,\hat{\mathbf i} + 2y \,\hat{\mathbf j}\) and \(\vecs\nabla g(x,y) = 4x \,\hat{\mathbf i} + 2y \,\hat{\mathbf j}\)

Using the Lagrange Multiplier equation, \[\vecs\nabla f(x, y) = \lambda\vecs\nabla g(x, y),\nonumber\]
we have: \[12x^2 \,\hat{\mathbf i} + 2y \,\hat{\mathbf j} = 4x\lambda \, \hat{\mathbf i} + 2y\lambda \,\hat{\mathbf j}\nonumber\]
giving us the system of equations: \[12x^2 = 4x\lambda, \quad 2y = 2y\lambda, \quad \text{and the constraint}\quad 2x^2 + y^2 = 1\nonumber\]
Rewriting the first two equations as zero-products (by moving to one-side and factoring), we get:
\[\begin{align*} 4x(3x - \lambda) &= 0 & \text{and} && 2y(1 - \lambda) &= 0 \\ x = 0 \quad \text{or}\quad \lambda &= 3x & & &y = 0 \quad \text{or}\quad \lambda &= 1 \end{align*}\]
Now we consider the combinations of these solutions to the above two equations and plug each of these into the constraint equation to solve for the corresponding Lagrange points.

The combination \(x = 0\) and \(y = 0\) produces a contradiction when placed in the constraint equation, since this point is not on the ellipse.

Taking the combination \(x = 0\) and \(\lambda = 1\), we put \(0\) in for \(x\) in the constraint and solve for \(y\), obtaining: \( y = \pm 1\). This gives us two Lagrange points: \( (0, 1) \) and \( (0, -1)\).

Taking the combination \(\lambda = 3x\) and \(y = 0\), we put \(0\) in for \(y\) in the constraint and solve for \(x\), obtaining: \( x = \pm \frac{\sqrt{2}}{2}\). This gives us two Lagrange points: \( \left(-\frac{\sqrt{2}}{2}, 0\right) \) and \( \left(\frac{\sqrt{2}}{2}, 0\right) \).

Taking the combination \(\lambda = 3x\) and \(\lambda = 1\), we substitute \(1\) into the first equation for \(\lambda\), giving us \( 1 = 3x\) so \(x = \frac{1}{3}\). Plugging this value in for \(x\) in the constraint equation and solving for \(y\), we obtain \(y = \pm \frac{\sqrt{7}}{3}\) which gives us the two Lagrange points: \( \left(\frac{1}{3}, -\frac{\sqrt{7}}{3}\right) \) and \( \left(\frac{1}{3}, \frac{\sqrt{7}}{3}\right) \).

Evaluating the function \(f\) at these Lagrange points, we find: \[\begin{align*} f(0, -1) &= 1 & f(0, 1) &= 1 \\ f\left(-\tfrac{\sqrt{2}}{2}, 0\right) &= \frac{-4(\sqrt{2})^3}{8} = -\sqrt{2} & f\left(\tfrac{\sqrt{2}}{2}, 0\right) &= \frac{4(\sqrt{2})^3}{8} = \sqrt{2} \\ f\left(\tfrac{1}{3}, -\tfrac{\sqrt{7}}{3}\right) &= \tfrac{25}{27} & f\left(\tfrac{1}{3}, \tfrac{\sqrt{7}}{3}\right) &= \tfrac{25}{27} \end{align*}\]

Comparing these values with where the corresponding Lagrange points lie on the constraint curve, we conclude the results stated in the answer above.

10) Objective function: \(f(x,y)=2x^2 +y^2\) Constraint: \( g(x,y)= x^2 +y^2 =1 \)

11) Objective function: \(f(x,y,z)=x+3y−z\) Constraint: \( x^2+y^2+z^2=4 \)

12) Objective function: \(f(x, y, z) = x + y + z\) Constraint: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

13) Objective function: \(f(x, y, z) = xyz\) Constraint: \(x^2+2y^2+3z^2=6\)

14) Objective function: \(f(x, y, z) = x^2 + y^2 + z^2\) Constraint: \(x^4+y^4+z^4=1\)

15) Objective function: \(f(x, y, z) = x^2 + y^2 + z^2\) Constraint: \(xyz=4\)

In exercises 16-21, use the method of Lagrange multipliers to find the requested extremum of the given function subject to the given constraint.

16) Maximize \(f(x,y) = \sqrt{6 - x^2 - y^2}\) subject to the constraint, \( x+y−2=0\).

17) Maximize \(f(x,y) = x^2 - y^2\) subject to the constraints, \( g(x,y)=y−x^2=0, \quad x>0,\quad y>0\).

18) Maximize \(U(x,y) = 8x^{4/5}y^{1/5}\) subject to the constraint, \( 4x+2y=12\).

19) Minimize \(f(x,y,z)=x^2+y^2+z^2\) subject to the constraint, \(x+y+z=1\).

20) Minimize \(f(x,y)=xy\) on the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

21) Maximize \(f(x,y,z)=2x+3y+5z\) on the sphere \(x^2+y^2+z^2=19\).

In exercises 22-23, use the method of Lagrange multipliers with two constraints.

22) Optimize \(f(x,y,z)=yz+xy\) subject to the constraints: \(xy=1, \quad y^2+z^2=1\).

23) Minimize \(f(x,y,z)=x^2+y^2+z^2\) when \(x+y+z=9\) and \(x+2y+3z=20\).

Use the method of Lagrange multipliers to solve the following applied problems.

24) A large container in the shape of a rectangular solid must have a volume of 480 m³. The bottom of the container costs $5/m² to construct whereas the top and sides cost $3/m² to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost.

25) A rectangular box without a top (a topless box) is to be made from 1212 ft² of cardboard. Find the maximum volume of such a box.

Solution:

26) Find the minimum distance from the parabola \(y=x^2\) to point \((0,3)\).

27) Find the point on the line \(y=2x+3\) that is closest to point \((4,2)\).

Solution:

28) A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is 1010 in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

29) Find the minimum distance from the plane \(x+y+z=1\) to point \((2,1,1)\).

Solution:

30) Find the point on the plane \(4x+3y+z=2 \) that is closest to the point \((1,−1,1)\).

31) [T] By investing \(x\) units of labor and \(y\) units of capital, a watch manufacturer can produce \(P(x,y)=50x^0.4y^0.6\) watches. Find the maximum number of watches that can be produced on a budget of $20,000 if labor costs $100/unit and capital costs $200/unit. Use a grapher like CalcPlot3D to sketch a contour plot of the function.

Solution:

32) A rectangular solid is contained within a tetrahedron with vertices at \((1,0,0),\,(0,1,0),\,(0,0,1)\), and the origin. The base of the box has dimensions \(x\) and \(y\), and the height of the box is \(z\). If the sum of \(x\), \(y\), and \(z\) is \(1\), find the dimensions that maximizes the volume of the rectangular solid.