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Partial Derivatives

2.10 Lagrange Multipliers

30 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1Past exam · M200 2010A
  1. Does the function f(x,y)=x2+y2f(x, y) = x^2 +y^2 have a maximum or a minimum on the curve xy=1xy = 1? Explain.

  2. Find all maxima and minima of f(x,y)f(x, y) on the curve xy=1xy = 1.

Answer

(a) ff does not have a maximum. It does have a minimum.

(b) The minima are at ±(1,1)\pm (1,1), where ff takes the value 22.

Full solution
Q2Stage 1

The surface SS is given by the equation g(x,y,z)=0g(x,y,z)=0. You are walking on SS measuring the function f(x,y,z)f(x,y,z) as you go. You are currently at the point (x0,y0,z0)(x_0,y_0,z_0) where ff takes its largest value on SS, and are walking in the direction d≠0\vd\ne\vZero. Because you are walking on SS, the vector d\vd is tangent to SS at (x0,y0,z0)(x_0,y_0,z_0).

  1. What is the directional derivative of ff at (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd? Do not use the method of Lagrange multipliers.

  2. What is the directional derivative of ff at (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd? This time use the method of Lagrange multipliers.

Hint

(a) The function ff decreases, or at least does not increase, as you leave (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd.
The function ff also decreases, or at least does not increase, as you leave (x0,y0,z0)(x_0,y_0,z_0) in the direction −d-\vd.

Answer

(a), (b) 00

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Find the maximum and minimum values of the function
f(x,y,z)=x+y−zf(x,y,z)=x+y-z on the sphere x2+y2+z2=1x^2+y^2+z^2=1.

Answer

The max is f=3f=\sqrt{3} and the min is f=−3f=-\sqrt{3}.

Full solution
Q4Stage 2

Find a, ba,\ b and cc so that the volume 4π3abc\frac{4\pi}{3} abc of an ellipsoid x2a2+y2b2+z2c2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 passing through the point (1,2,1)(1,2,1) is as small as possible.

Hint

The ellipsoid x2a2+y2b2+z2c2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 passes through the point (1,2,1)(1,2,1) if and only if 1a2+4b2+1c2=1\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=1.

Answer

a=c=3, b=23a=c=\sqrt{3},\ b=2\sqrt{3}.

Full solution
Q5Stage 2Past exam · M200 2005D

Use the Method of Lagrange Multipliers to find the minimum value of z=x2+y2z = x^2 + y^2 subject to x2y=1x^2 y = 1. At which point or points does the minimum occur?

Answer

The minimum value is 213+2−23=3223=3432^{\frac{1}{3}} + 2^{-\frac{2}{3}} =\frac{3}{2}\sqrt[3]{2} =\frac{3}{\sqrt[3]{4}} at (±216 , 2−13)\big(\pm 2^{\frac{1}{6}}\,,\, 2^{-\frac{1}{3}}\big).

Full solution
Q6Stage 2Past exam · M200 2006A

Use the Method of Lagrange Multipliers to find the radius of the base and the height of a right circular cylinder of maximum volume which can be fit inside the unit sphere x2+y2+z2=1x^2 + y^2 + z^2 = 1.

Answer

radius=23\text{radius}=\sqrt{\frac{2}{3}} and height=23\text{height}=\frac{2}{\sqrt{3}}.

Full solution
Q7Stage 2Past exam · M200 2006D

Use the method of Lagrange Multipliers to find the maximum and minimum values of

f(x,y)=xy\begin{equation*} f(x, y) = xy \end{equation*}

subject to the constraint

x2+2y2=1.\begin{equation*} x^2 + 2y^2 = 1. \end{equation*}
Answer

The maximum and minimum values of ff are 122\frac{1}{2\sqrt{2}} and −122-\frac{1}{2\sqrt{2}}, respectively.

Full solution
Q8Stage 2Past exam · M200 2008A

Find the maximum and minimum values of f(x,y)=x2+y2f(x,y) = x^2 + y^2 subject to the constraint x4+y4=1x^4 + y^4 = 1.

Answer

min=1=1, max=2=\sqrt{2}.

Full solution
Q9Stage 2Past exam · M200 2008D

Use Lagrange multipliers to find the points on the sphere z2+x2+y2−2y−10=0z^2 + x^2 + y^2 - 2y - 10 = 0 closest to and farthest from the point (1,−2,1)(1, -2, 1).

Answer

(1,−2,1)(1,-2,1) is the closest point. (−1,4,−1)(-1,4,-1) is the farthest point.

Full solution
Q10Stage 2Past exam · M200 2009A

Use Lagrange multipliers to find the maximum and minimum values of the function f(x,y,z)=x2+y2−120z2f(x,y,z) = x^2 + y^2 -\frac{1}{20} z^2 on the curve of intersection of the plane x+2y+z=10x + 2y + z = 10 and the paraboloid x2+y2−z=0x^2 + y^2 - z = 0.

Answer

The maximum is 55 and the minimum is 00.

Full solution
Q11Stage 2Past exam · M200 2010D

Find the point P=(x,y,z)P = (x, y, z) (with xx, yy and z>0z> 0) on the surface x3y2z=63x^3 y^2 z = 6 \sqrt{3} that is closest to the origin.

Answer

(3 , 2 , 1)\big(\sqrt{3}\,,\,\sqrt{2}\,,\,1\big)

Full solution
Q12Stage 2Past exam · M200 2011A

Find the maximum value of f(x,y,z)=xyzf (x, y, z) = xyz on the ellipsoid

g(x,y,z)=x2+xy+y2+3z2=9\begin{equation*} g(x, y, z) = x^2 + xy + y^2 + 3z^2 = 9 \end{equation*}

Specify all points at which this maximum value occurs.

Answer

The maximum is 66 and is achieved at (6 , −6 , −1)\big(\sqrt{6}\,,\,-\sqrt{6}\,,\,-1\big) and (−6 , 6 , −1)\big(-\sqrt{6}\,,\,\sqrt{6}\,,\,-1\big).

Full solution
Q13Stage 2Past exam · M200 2011D

Find the radius of the largest sphere centred at the origin that can be inscribed inside (that is, enclosed inside) the ellipsoid

2(x+1)2+y2+2(z−1)2=8\begin{equation*} 2(x+1)^2 + y^2 + 2(z-1)^2 =8 \end{equation*}
Answer

6−42≈0.59\sqrt{6-4\sqrt{2}}\approx 0.59

Full solution
Q14Stage 2Past exam · M200 2012a

Let CC be the intersection of the plane x+y+z=2x + y + z = 2 and the sphere x2+y2+z2=2x^2 + y^2 + z^2 = 2.

  1. Use Lagrange multipliers to find the maximum value of f(x,y,z)=zf(x, y, z) = z on CC.

  2. What are the coordinates of the lowest point on CC?

Answer

(a) 43\frac{4}{3} (b) (1,1)(1,1)

Full solution
Q15Stage 2Past exam · M200 2012D
  1. Use Lagrange multipliers to find the extreme values of

    f(x,y,z)=(x−2)2+(y+2)2+(z−4)2\begin{equation*} f (x, y, z) = (x - 2)^2 + (y + 2)^2 + (z - 4)^2 \end{equation*}

    on the sphere x2+y2+z2=6x^2 + y^2 + z^2 = 6.

  2. Find the point on the sphere x2+y2+z2=6x^2 + y^2 + z^2 = 6 that is farthest from the point (2,−2,4)(2, -2, 4).

Answer

(a) The min is 66 and the max is 5454. (b) (−1,1,−2)(-1,1,-2)

Full solution
Q16Stage 2Past exam · M200 2013D
  1. Find the minimum of the function

    f(x,y,z)=(x−2)2+(y−1)2+z2\begin{equation*} f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2 \end{equation*}

    subject to the constraint x2+y2+z2=1x^2 + y^2 + z^2 = 1, using the method of Lagrange multipliers.

  2. Give a geometric interpretation of this problem.

Answer

(a) (5−1)2=6−25\big(\sqrt{5}-1\big)^2=6-2\sqrt{5}

(b) The minimum of ff subject to the constraint x2+y2+z2=1x^2+y^2+z^2=1 is the square of the distance from (2,1,0)(2,1,0) to the point on the sphere x2+y2+z2=1x^2+y^2+z^2=1 that is nearest (2,1,0)(2,1,0).

Full solution
Q17Stage 2Past exam · M200 2015D

Use Lagrange multipliers to find the minimum and maximum values of (x+z)ey(x + z)e^y subject to x2+y2+z2=6x^2 + y^2 + z^2 = 6.

Answer

The maximum value is 2e22e^2 and the minimum value is −2e2-2 e^2.

Full solution
Q18Stage 2Past exam · M200 2004A

Find the points on the ellipse 2x2+4xy+5y2=302x^2 + 4xy + 5y^2 = 30 which are closest to and farthest from the origin.

Answer

The farthest points are ±6(−2,1)\pm\sqrt{6}(-2,1). The nearest points are ±(1,2)\pm(1,2).

Full solution
Q19Stage 2

Find the ends of the major and minor axes of the ellipse 3x2−2xy+3y2=43x^2-2xy+3y^2=4.

Hint

The ends of the major axes are the points on the ellipse which are farthest from the origin. The ends of the minor axes are the points on the ellipse which are closest to the origin.

Answer

The ends of the minor axes are ±(12,−12)\pm\big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\big). The ends of the major axes are ±(1,1)\pm(1,1).

Full solution
Q20Stage 2Past exam · M200 2003D

A closed rectangular box with a volume of 96 cubic meters is to be constructed of two materials. The material for the top costs twice as much per square meter as that for the sides and bottom. Use the method of Lagrange multipliers to find the dimensions of the least expensive box.

Answer

x=y=4, z=6 metersx=y=4,\ z=6\ \text{meters}

Full solution
Q21Stage 2Past exam · M200 2003A

Consider the unit sphere

S={ (x,y,z) ∣ x2+y2+z2=1 }\begin{equation*} S=\Set{(x,y,z)}{x^2+y^2+z^2=1} \end{equation*}

in R3\bbbr^3. Assume that the temperature at a point (x,y,z)(x,y,z) of SS is

T(x,y,z)=40xy2z\begin{equation*} T(x,y,z)=40xy^2z \end{equation*}

Find the hottest and coldest temperatures on SS.

Answer

The hottest temperature is +5+5 and the coldest temperature is −5-5.

Full solution
Q22Stage 2Past exam · M200 2002A

Find the dimensions of the box of maximum volume which has its faces parallel to the coordinate planes and which is contained inside the region 0≤z≤48−4x2−3y20\le z\le 48-4x^2-3y^2.

Figure from prob_s2.10, line 1458

Figure from prob_s2.10, line 1458

Answer

23×4×242\sqrt{3} \times 4\times 24

Full solution
Q23Stage 2Past exam · M200 2001D

A rectangular bin is to be made of a wooden base and heavy cardboard with no top. If wood is three times more expensive than cardboard, find the dimensions of the cheapest bin which has a volume of 12m312{\rm m}^3.

Answer

2m×2m×3m2{\rm m}\times 2{\rm m}\times 3{\rm m}

Full solution
Q24Stage 2Past exam · M200 2000D

A closed rectangular box having a volume of 44 cubic metres is to be built with material that costs $8 per square metre for the sides but $12 per square metre for the top and bottom. Find the least expensive dimensions for the box.

Answer

233m×233m×32/3m\frac{2}{\sqrt[3]{3}}{\rm m}\times\frac{2}{\sqrt[3]{3}}{\rm m} \times 3^{2/3}{\rm m}

Full solution
Q25Stage 2Past exam · M200 2000A

Suppose that aa, bb, cc are all greater than zero and let DD be the pyramid bounded by the plane ax+by+cz=1ax+by+cz=1 and the 3 coordinate planes. Use the method of Lagrange multipliers to find the largest possible volume of DD if the plane ax+by+cz=1ax + by + cz = 1 is required to pass through the point (1,2,3)(1, 2, 3). (The volume of a pyramid is equal to one-third of the area of its base times the height.)

Answer

a=13a=\frac{1}{3}, b=16b=\frac{1}{6}, c=19c=\frac{1}{9}, max volume=27= 27

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q26Stage 3Past exam · M200 2009D

Use Lagrange multipliers to find the minimum distance from the origin to all points on the intersection of the curves

g(x,y,z)=x−z−4=0and h(x,y,z)=x+y+z−3=0\begin{align*} g(x,y,z) &= x-z-4=0 \\ \text{and } h(x,y,z) &= x+y+z-3=0 \end{align*}
Answer

11\sqrt{11}

Full solution
Q27Stage 3Past exam · M200 2014A

Find the largest and smallest values of

f(x,y,z)=6x+y2+xz\begin{equation*} f(x,y,z) = 6x + y^2 + xz \end{equation*}

on the sphere x2+y2+z2=36x^2 + y^2 + z^2 = 36. Determine all points at which these values occur.

Answer

The min is −273-27\sqrt{3} at (−33,0,3)(-3\sqrt{3},0,3) and the max is 4848 at (4,±4,2)(4,\pm 4,2).

Full solution
Q28Stage 3Past exam · M200 2016D

The temperature in the plane is given by T(x,y)=ey(x2+y2)T(x,y) = e^y\big(x^2+y^2\big).

    1. Give the system of equations that must be solved in order to find the warmest and coolest point on the circle x2+y2=100x^2+y^2=100 by the method of Lagrange multipliers.

    2. Find the warmest and coolest points on the circle by solving that system.

    1. Give the system of equations that must be solved in order to find the critical points of T(x,y)T(x,y).

    2. Find the critical points by solving that system.

  1. Find the coolest point on the solid disc x2+y2≤100x^2+y^2\le 100.

Answer

(a) (i)

2x ey=λ(2x)ey(x2+y2+2y)=λ(2y)x2+y2=100\begin{align*} 2x\,e^y &=\la (2x) \\ e^y\big(x^2+y^2+2y\big) &=\la (2y) \\ x^2+y^2&=100 \end{align*}

(a) (ii) The warmest point is (0,10)(0,10) and the coolest point is (0,−10)(0,-10).

(b) (i)

2x ey=0ey(x2+y2+2y)=0\begin{align*} 2x\,e^y &=0 \\ e^y\big(x^2+y^2+2y\big) &=0 \end{align*}

(b) (ii) (0,0)(0,0) and (0,−2)(0,-2)

(c) (0,0)(0,0)

Full solution
Q29Stage 3Past exam · M200 2002D
  1. By finding the points of tangency, determine the values of cc for which x+y+z=cx+y+z=c is a tangent plane to the surface 4x2+4y2+z2=964x^2+4y^2+z^2=96.

  2. Use the method of Lagrange Multipliers to determine the absolute maximum and minimum values of the function f(x,y,z)=x+y+zf(x,y,z)=x+y+z along the surface g(x,y,z)=4x2+4y2+z2=96g(x,y,z)=4x^2+4y^2+z^2=96.

  3. Why do you get the same answers in (a) and (b)?

Answer

(a) c=±12c=\pm 12 (b) ±12\pm 12

(c) The level surfaces of x+y+zx+y+z are planes with equation of the form x+y+z=cx+y+z=c. To find the largest (smallest) value of x+y+zx+y+z on 4x2+4y2+z2=964x^2+4y^2+z^2=96 we keep increasing (decreasing) cc until we get to the largest (smallest) value of cc for which the plane x+y+z=cx+y+z=c intersects 4x2+4y2+z2=964x^2+4y^2+z^2=96. For this value of cc, x+y+z=cx+y+z=c is tangent to 4x2+4y2+z2=964x^2+4y^2+z^2=96.

Full solution
Q30Stage 3

Let f(x,y)f(x,y) have continuous partial derivatives. Consider the problem of finding local minima and maxima of f(x,y)f(x,y) on the curve xy=1xy=1.

  • Define g(x,y)=xy−1g(x,y) = xy -1. According to the method of Lagrange multipliers, if (x,y)(x,y) is a local minimum or maximum of f(x,y)f(x,y) on the curve xy=1xy=1, then there is a real number λ\la such that

    ∇f(x,y)=λ∇g(x,y),g(x,y)=0\begin{equation*} \vnabla f(x,y) =\la \vnabla g(x,y),\quad g(x,y)=0 \tag{E1}\end{equation*}
  • On the curve xy=1xy=1, we have y=1xy=\frac{1}{x} and f(x,y)=f(x,1x)f(x,y) =f\big(x,\frac{1}{x}\big). Define F(x)=f(x,1x)F(x)=f\big(x,\frac{1}{x}\big). If x≠0x\ne 0 is a local minimum or maximum of F(x)F(x), we have that

    F′(x)=0\begin{equation*} F'(x)=0 \tag{E2}\end{equation*}

Show that (E1) is equivalent to (E2), in the sense that

there is a λ such that (x,y,λ) obeys (E1)if and only ifx≠0 obeys (E2) and y=1 ⁣/x.\begin{align*} &\text{there is a }\la\text{ such that }(x,y,\la)\text{ obeys (E1)}\\ &\hskip-0.5in\text{if and only if}\\ &x\ne 0\text{ obeys (E2) and }y=\nicefrac{1}{x}\text{.} \end{align*}
Answer

See the solution.

Full solution

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.