Applications: Functions of Several Variables

The cost depends on two inputs at once, the side of the base and the height, so it is a function of two variables. It is built face by face: each face's area times the price of its plywood. Part (a) substitutes both inputs for each design. In part (b) one input is fixed, which leaves a function of the side alone, and setting it equal to the budget gives a quadratic equation.

1. The base and the lid are two squares of area x², so together they cost 12 × 2x² = 24x² dollars. The four sides are rectangles of area xy, so they cost 8 × 4xy = 32xy dollars. Hence C(x, y) = 24x² + 32xy.
2. Both designs hold what the question says: 1.5 × 1.5 × 2 = 4.5 and 3 × 3 × 0.5 = 4.5 cubic meters. (a) For design A, C(1.5, 2) = 24 × 2.25 + 32 × 3 = 54 + 96 = 150 dollars.
3. For design B, C(3, 0.5) = 24 × 9 + 32 × 1.5 = 216 + 48 = 264 dollars. The two designs hold the same amount, but B costs $114 more, because its large base and lid use the more expensive plywood.
4. (b) With the height fixed at y = 2, the cost is a function of x alone: C(x, 2) = 24x² + 64x. Setting it equal to the budget gives 24x² + 64x = 224, and dividing by 8 gives 3x² + 8x − 28 = 0.
5. This factorizes as (3x + 14)(x − 2) = 0, so x = 2 or x = −143. A side cannot be negative, so the base is 2 meters square. Check: 24 × 4 + 32 × 2 × 2 = 96 + 128 = 224 dollars.

answer(a) C(x, y) = 24x² + 32xy; design A costs $150 and design B costs $264; (b) a base 2 meters square

techniqueFunctions of Two Variables · Level Curves

The range depends on two inputs, the speed and the angle, so comparing two kicks means substituting both inputs each time: a faster kick does not always go farther. Holding the speed fixed takes a cross-section of the function, which leaves the range as a function of the angle alone. Setting it equal to the distance required gives an equation in sin 2θ, which can have more than one solution between 0^° and 90^°.

1. (a) For the first kick, sin 90^° = 1, so R(20, 45^°) = 400 × 110 = 40 meters.
2. For the second kick, sin 30^° = 0.5, so R(25, 15^°) = 625 × 0.510 = 31.25 meters. The first kick travels 8.75 meters farther: the second kick's greater speed does not make up for its low angle.
3. (b) Hold the speed at v = 20. The range is then a function of the angle alone, R(20, θ) = 400 sin 2θ10 = 40sin 2θ, and it must equal 20, so sin 2θ = 0.5.
4. Because θ lies between 0^° and 90^°, 2θ lies between 0^° and 180^°, where the sine is 0.5 twice: 2θ = 30^° or 2θ = 150^°. So θ = 15^° or θ = 75^°.
5. Check: sin 30^° = sin 150^° = 0.5, so both angles give 40 × 0.5 = 20 meters. The low kick at 15^° reaches the teammate sooner; the high kick at 75^° spends longer in the air.

answer(a) the kick at 20 meters per second and 45^°, which travels 40 meters against 31.25 meters; (b) θ = 15^° or θ = 75^°

techniqueFunctions of Two Variables · Surfaces in Three Dimensions

The roof is the graph of a function of two variables, a surface above the floor. A surface is understood through its cross-sections: holding y constant gives the shape of the roof along an east–west line, and holding x constant gives its shape along a north–south line. Each cross-section is a function of one variable, which can be read like any graph.

1. (a) The middle of the east wall is (10, 0), where z = 12 + 100 − 050 = 14 meters; the middle of the west wall, (−10, 0), gives the same. The middle of the north wall is (0, 10), where z = 12 + 0 − 10050 = 10 meters, and the south wall gives the same.
2. At each corner, such as (10, 10), x² = y² = 100, so z = 12 + 0 = 12 meters, the same height as above the center (0, 0).
3. Along the east–west center line y = 0, the roof is z = 12 + x²50, a parabola that opens upward, so the center is its lowest point. Along the north–south line x = 0, the roof is z = 12 − y²50, a parabola that opens downward, so the center is its highest point. A surface that is lowest in one direction and highest in the other at the same point has a saddle point there.
4. (b) Hold x = 5 for the cross-section under the truss: z = 12 + 25 − y²50 = 12.5 − y²50. This is a parabola that opens downward, the same shape as the cross-section x = 0 raised by 0.5 meters.
5. The truss is highest where y = 0, at 12.5 meters. Its ends are at the south and north walls, y = ± 10, where z = 12.5 − 10050 = 10.5 meters. Check: at the north end, (5, 10), the formula gives 12 + 25 − 10050 = 12 − 1.5 = 10.5 meters.

answer(a) 14 meters above the middles of the east and west walls, 10 meters above the middles of the north and south walls, and 12 meters above each corner; along y = 0 the roof is a parabola that opens upward and along x = 0 one that opens downward, so the center is a saddle point; (b) 12.5 meters at its highest point, above y = 0, and 10.5 meters at each end

techniqueSurfaces in Three Dimensions · Functions of Two Variables

The displacement depends on where the point is and on when it is observed, so it is a function of two variables. Its two kinds of cross-section answer the two parts. Holding the time fixed gives the shape of the whole string at one instant, like a photograph. Holding the position fixed gives the motion of one point over time, like watching a single bead on the string.

1. (a) Hold the time fixed at s = 1600. Then cos(200π s) = cosπ3 = 0.5, so the displacement along the string is u = 2 × 0.5 × sinπ x60 = sinπ x60 millimeters, for 0 ≤ x ≤ 60.
2. This is one arch of a sine curve. It is 0 at both fixed ends, x = 0 and x = 60, and greatest where π x60 = π2, that is at the middle, x = 30, where the displacement is 1 millimeter.
3. (b) Now hold the position fixed at x = 10. There sinπ6 = 0.5, so u = 2 × 0.5 × cos(200π s) = cos(200π s) millimeters. The point moves up and down in simple harmonic motion with an amplitude of 1 millimeter.
4. One full oscillation takes the time for 200π s to grow by 2π, which is 2π200π = 0.01 seconds. So the period is 0.01 seconds, and there are 100 oscillations in one second.
5. In each oscillation the point goes from 1 millimeter above its rest position to 1 millimeter below and back again, a distance of 4 × 1 = 4 millimeters. In one second it travels 100 × 4 = 400 millimeters, which is 40 centimeters. Check: the middle of the string, with an amplitude of 2 millimeters, travels twice as far, 800 millimeters.

answer(a) u = sinπ x60 millimeters, one arch of a sine curve, greatest at the middle of the string, where it is 1 millimeter; (b) an amplitude of 1 millimeter and a period of 0.01 seconds, so the point travels 400 millimeters in one second

techniqueSurfaces in Three Dimensions · Functions of Two Variables

A contour on a map is a level curve: it joins every point where the height function takes one value. Its equation comes from setting h(x, y) equal to that height. Where contours lie close together the ground is steep, and the average gradient of a path is the drop in height divided by the distance along the ground, both in meters.

1. (a) At the hut, x² + 4y² = 0.36 + 4 × 0.16 = 1, so its height is h = 450 − 200 × 1 = 250 meters.
2. (b) The contour through the hut joins every point at a height of 250 meters: 450 − 200(x² + 4y²) = 250, which gives x² + 4y² = 1. This is an ellipse centered on the summit, and every contour of this hill is an ellipse of the same shape.
3. On the path due east, y = 0, so x² = 1 and the path meets the contour 1 kilometer from the summit. On the path due north, x = 0, so 4y² = 1 and the path meets the contour 0.5 kilometers from the summit.
4. Both paths drop 450 − 250 = 200 meters to the contour. The average gradient is 2001000 = 0.2 due east and 200500 = 0.4 due north, so the path due north is on average twice as steep.
5. Check: the point (0, 0.5) gives h = 450 − 200 × 4 × 0.25 = 250 meters, as it should. On the map the contours lie closer together to the north of the summit, which is what steeper ground looks like.

answer(a) 250 meters; (b) x² + 4y² = 1, an ellipse that crosses the east path 1 kilometer and the north path 0.5 kilometers from the summit; the average gradients are 0.2 due east and 0.4 due north, so the north path is twice as steep

techniqueLevel Curves · Functions of Two Variables

An isobar is a level curve of the pressure: it joins the points where the pressure takes one value. Setting p(x, y) equal to a constant gives its equation, and for this model every such equation is a straight line. Lines with the same gradient are parallel, and the distance between two parallel lines is measured along a line at right angles to both.

1. (a) At the town, p(100, 200) = 1012 + 0.02 × 100 − 0.015 × 200 = 1012 + 2 − 3 = 1011 hectopascals.
2. The isobar through the town joins the points where p = 1011: 0.02x − 0.015y = −1. Multiplying by 200 gives 4x − 3y = −200. Check: 4 × 100 − 3 × 200 = −200.
3. (b) In the same way, the isobar p = c is the line 4x − 3y = 200(c − 1012). Every one of these lines has the gradient 43, so the isobars are parallel straight lines. The 1012 isobar is 4x − 3y = 0 and the 1008 isobar is 4x − 3y = −800.
4. The distance between the parallel lines 4x − 3y = 0 and 4x − 3y = −800 is the difference of the constants divided by √4² + 3² = 5, which is 8005 = 160 kilometers.
5. Check: the point of the 1008 isobar nearest the station is (−128, 96), since 4 × (−128) − 3 × 96 = −512 − 288 = −800, and its distance from the station is √16384 + 9216 = √25600 = 160 kilometers. The pressure falls by 4 hectopascals over those 160 kilometers, which is 2.5 hectopascals per 100 kilometers.

answer(a) 1011 hectopascals, on the isobar 4x − 3y = −200; (b) 160 kilometers

techniqueLevel Curves · Functions of Two Variables

An isoquant is a level curve of the output: every mix of workers and machines on it makes the same number of units. Holding the output at its present value ties K to L, so the weekly cost along the isoquant becomes a function of L alone. Its least value is where its derivative is zero, confirmed by the second derivative.

1. (a) At present, Q(16, 25) = 20√16 × 25 = 20√400 = 20 × 20 = 400 units a week.
2. The isoquant is the level curve Q = 400: 20√LK = 400, so √LK = 20 and LK = 400. Every mix on it, such as 20 workers with 20 machines, makes the same 400 units.
3. (b) The weekly cost is C = 1200L + 300K dollars. On the isoquant K = 400L, so along it C(L) = 1200L + 120000L.
4. Then dCdL = 1200 − 120000L², which is 0 when L² = 100, so L = 10, since a number of workers is positive. The second derivative, 240000L³, is positive there, so this is a minimum, and K = 40010 = 40 machines.
5. The cheapest mix costs 1200 × 10 + 300 × 40 = 12000 + 12000 = $24000 a week. The present mix costs 1200 × 16 + 300 × 25 = 19200 + 7500 = $26700 a week, so the cheapest mix saves $2700 a week for the same output. Check: Q(10, 40) = 20√400 = 400 units.

answer(a) 400 units a week, on the isoquant LK = 400; (b) 10 workers and 40 machines, at $24000 a week against $26700 at present, a saving of $2700 a week

techniqueLevel Curves · Functions of Two Variables

A limit at a point of the plane must be the same along every path into that point. Straight lines through the point are the simplest paths to test: if two of them give different values, there is no limit. Here each straight line through (0, 0) keeps the ratio of the two flows fixed, and so it keeps the temperature fixed too.

1. (a) T(6, 3) = 60 × 6 + 15 × 36 + 3 = 360 + 459 = 4059 = 45^°C.
2. (b) Close the tap keeping the cold flow a fixed multiple of the hot flow, y = mx with m ≥ 0. Along that line T = 60x + 15mxx + mx = 60 + 15m1 + m. The x cancels, so the temperature stays the same all the way in to (0, 0).
3. Along y = 0.5x, the mix in (a), the value is 60 + 7.51.5 = 45^°C; along y = x it is 752 = 37.5^°C. Two paths into (0, 0) give different values, so T has no limit there, and no value given to T(0, 0) could make T continuous there.
4. For 42^°C, solve 60 + 15m1 + m = 42: 60 + 15m = 42 + 42m, so 27m = 18 and m = 23. The cold flow is two thirds of the hot flow, a ratio of hot to cold of 3 : 2.
5. Check: with 3 liters per minute of hot water and 2 of cold, T = 180 + 305 = 42^°C, and halving both flows to 1.5 and 1 gives 90 + 152.5 = 42^°C again.

answer(a) 45^°C; (b) along y = 0.5x the temperature stays at 45^°C and along y = x it stays at 37.5^°C, so there is no limit; a ratio of hot to cold of 3 : 2 keeps it at 42^°C

techniqueLimits of Functions of Two Variables · Level Curves · Continuity in Two Variables

Substituting (10, 10) gives 00, which says only that substitution fails. To find the limit, write the second difference as a multiple of the first, y = x(1 + u). Then D becomes x times a quantity that depends on u alone, and a standard one-variable limit settles that quantity for every path at once. A function is continuous at a point when its value there equals its limit there.

1. (a) D(40, 10) = 40 − 10ln 40 − ln 10 = 30ln 4 = 301.3863 ≈ 21.6^°C. The ordinary average is 40 + 102 = 25^°C, so the log mean is smaller, by about 3.4^°C.
2. (b) Write y = x(1 + u), where u = yx − 1. Then x − y = −xu and ln x − ln y = −ln(1 + u), so D = x × uln(1 + u) whenever u ≠ 0.
3. As (x, y) approaches (10, 10) along any path, x approaches 10 and u = yx − 1 approaches 1010 − 1 = 0. The standard limit ln(1 + u)u → 1 as u → 0 gives uln(1 + u) → 1.
4. So D approaches 10 × 1 = 10 whichever way (x, y) approaches (10, 10): the limit is 10^°C. Defining D(10, 10) = 10, the common difference, makes D continuous there; in the same way D(a, a) = a for every positive a.
5. Check: D(10, 10.2) = −0.2ln 10 − ln 10.2 = −0.2−0.0198 ≈ 10.1, just above 10, as it should be for differences at or a little above 10^°C.

answer(a) 30ln 4 ≈ 21.6^°C, about 3.4^°C below the ordinary average of 25^°C; (b) the limit is 10^°C, so D(10, 10) = 10 makes D continuous there

techniqueContinuity in Two Variables · Limits of Functions of Two Variables

Testing paths can prove that a limit does not exist, but it cannot prove that one does, because there are infinitely many paths. A bound can: if 0 ≤ R ≤ B near the point and B approaches 0 however the point is approached, then R is squeezed to 0 along every path at once. The bound here comes from the fact that a square is never negative.

1. (a) R(6, 3) = 6 × 36 + 3 = 189 = 2 ohms, less than either rheostat on its own, as a pair in parallel always is.
2. (b) For the bound, start from (x − y)² ≥ 0, which expands to x² + y² ≥ 2xy. Adding 2xy to both sides gives (x + y)² ≥ 4xy.
3. Dividing both sides by 4(x + y), which is positive everywhere except at (0, 0), gives xyx + y ≤ x + y4. So 0 ≤ R(x, y) ≤ x + y4 for every setting.
4. As (x, y) approaches (0, 0) along any path, x + y4 approaches 0, and R is squeezed between 0 and it. So the limit of R is 0 ohms, and defining R(0, 0) = 0, which is what the circuit gives when both rheostats are turned right down, makes R continuous there.
5. Check: with both rheostats below 0.1 ohms, the bound gives R < 0.1 + 0.14 = 0.05 ohms. The settings 0.08 and 0.02 ohms give R = 0.00160.1 = 0.016 ohms, below that bound.

answer(a) 2 ohms; (b) R ≤ x + y4, so R approaches 0 ohms along every path, and R(0, 0) = 0 makes R continuous at (0, 0)

techniqueLimits of Functions of Two Variables · Continuity in Two Variables