Solutions to Homework 7

14.1

#5 To maximize −x^{2} − y^{2} it is enough to minimize both x^{2} and y^{2}, so that the maximum is achieved when x = y = 0, which is in the inside of the square, and there z = 0. To minimize −x^{2} − y^{2} it is enough to maximize both x^{2} and y^{2}, so that the minimum is achieved if x = −1, 1 and y = −1, 1, which are the corner points of the boundary, and there z = −2.

#6 At the maximum value for f we get f_{x} = −2x − B = 0, so for the maximum to have x = −2, we need B = 4. Similarly f_{y} = −2y − C = 0 for y = 1 implies that C = −2. Now in order to get f(−2, 1) = 15 we finally need A = 10. Also to check that we have a maximum, note that D = (−2)(−2) − 0^{2} > 0 and f_{xx} < 0.

#10 [5]

a) To find the critical points, let f_{x} = 8y − (x + y)^{3} = 0, and f_{y} = 8x − (x + y)^{3} = 0. Subtracting the equations from each other we get that 8y − 8x = 0 or x = y. Plugging this back into one of the equations we get that 0 = 8x − (2x)^{3} = 8x(1 − x^{2}), so that x = 0, 1, −1 and therefore the critical points are (-1,-1), (0,0) and (1,1).

b) f_{xx} = f_{yy} = −3(x+y)^{2} and f_{xy} = 8−3(x+y)^{2}, so that D = 9(x+y)^{4} −(8−(x+ y)^{2})^{2} and for (0,0) we get D = −64 < 0, so that we get a saddle point, whereas for the other two points we get D = 128 > 0 and f_{xx} = −12 < 0, so that (-1,-1) and (1,1) are local maxima.

#11 P_{x} = −6x − 4 + 2y and P_{y} = 2x − 10y + 48. Setting those zero we get the critical point (1,5). Next P_{xx} = −6, P_{xy} = 2 and P_{yy} = −10, so that D = 56 > 0 and thus (1,5) is a local maximum.

#16 f_{x} = y +1/x and f_{y} = x+2y, so setting f_{y} = 0 we get that x = −2y and plugging this into f_{x} = 0 and factoring we get that 0 = \frac{1}{y} (y^{2} − 1/2). Thus y = +\frac{\sqrt{2}}{2}, −\frac{\sqrt{2}}{2}

and the solutions are (\sqrt{2}, \frac{\sqrt{2}}{2}) and (−\sqrt{2}, −\frac{\sqrt{2}}{2}), but x > 0 so that (\sqrt{2}, \frac{\sqrt{2}}{2}) is the only critical point. Since we can compute, that D = −\frac{2}{x^{2}} − 1 we get this point is a saddle point.

#19 [5] It is a global (and thus local) maximum, since x^{6} ≥ 0 and y^{6} ≥ 0 so that f(x, y) ≥ 0 = f(0, 0).

#24 The correct diagram would look almost look like figure 14.8 on pg 188, with the only difference being that you rotate the contour line from (-3,-3) to (3,3) by 45^{o}, so that it agrees with the y-axis. Consecutively the negative contours will be more squashed and the positive contours will be wider. Also certainly the intersection of the two contours with z = 0 is the point (a, b) rather then the origin, and it will be a saddle point. A possible equation for a function with these contours is z = (x − a)^{2} + (x − a)(y − b).

14.2

#2 The total revenue is R = pq = 60q − 0.04q^{2} and since q = q_{1} + q_{2}, we get that the profit is

P = R − C_{1} − C_{2} = −13.7 + 60q_{1} + 60q_{2} − 0.07q_{1}^{2} − 0.08q_{2}^{2} − 0.08q_{1}q_{2}.

Next ∂P/∂q_{1} = 60 − 0.14q_{1} − 0.08q_{2} = 0 and ∂P/∂q_{2} = 60 − 0.16q_{2} − 0.08q_{1} = 0 gives us after solving this system of linear equations the critical point (q_{1}, q_{2}) = (300, 225). Next D = (−0.14)(−0.16) − (−.08)^{2} > 0, so that our point is a local maximum. Since our function is a bowl opening downwards we get a global maximum. This can also seen by checking the boundaries q_{1} = 0 and q_{2} = 0 and it will turn out that the values taken on these curves are smaller.

#7 Here C = k(q_{a} + q_{c}) for some constant k, so that

P = q_{c}p_{c} + q_{a}p_{a} − k(q_{a} + q_{c}) = rp_{c}^{−3} + sp_{a}^{−1} − krp_{c}^{−4} − ksp_{a}^{−2}.

So ∂P/∂p_{a} = −sp_{a}^{−2} + 2ksp_{a}^{−3} and ∂P/∂p_{c} = −3rp_{c}^{−4} + 4ksp_{a}^{−5}. Setting those equal to 0 r and s drop out and we get −p_{a} + 2k = 0 and −3p_{c} + 4k = 0, so that subtracting twice the first equation from the second and solving we get that p_{a}/p_{c} = 3/2.

#9 [5] 512 = V = whl and C = 2hw+2hl+4wl is the cost function we want to minimize. Solving V for h and replacing all h’s in C we get that C = 1024/l+1024/w +4wl. Next C_{w} = −1024/w^{2} + 4l = 0 and C_{l} = −1024/l^{2} + 4w = 0. Solving both equations for 1024 and setting them equal we get that 4lw^{2} = 4l^{2}w or 4lw(l−w) = 0 and since l and w are not zero (V = 512) we can conclude that l = w and using C_{w} we get that w = 256^{1/3} = l and then using V we get that h = 2 ∗ 256^{1/3}. This is a local minimum, since D = 16 − 4 > 0 and C_{ww} > 0. Since C increases without bound as w, h → 0 or ∞ this must be a global minimum.

#12

a) Points which are directly below each other share the same x-coordinate, therefore the point on the least squares line directly above or below x_{i} has a y-coordinate of b + mx_{i}.

b) The square of the distance between the two points is d^{2} = \sqrt{(x_{i} − x_{i})^{2} + (y_{i} − (b + mx_{i}))^{2}}^{2} = (y_{i} − (b + mx_{i}))^{2}.

c) f_{b} = ∑ 2(y_{i} − (b + mx_{i}))(−1) = −2 ∑(y_{i} − (b + mx_{i})), and f_{m} = ∑ 2(y_{i} − (b + mx_{i}))(−x_{i}) = −2 ∑(y_{i} − (b + mx_{i}))x_{i}.

d) f_{b} = −2((∑ y_{i}) − (∑ b) − (m ∑ x_{i})) = −2SX + 2bn + mSX = 0 and f_{m} = −2((∑ y_{i}x_{i}) − (b ∑ x_{i}) − (m ∑ x_{i}^{2})) = −2SY X + 2bSX + 2mSXX = 0. Solving these equations for b and m we get that b = (SXX SY − SX SY X)/(nSXX − (SX)^{2}) and m = (nSY X − SX SY)/(nSXX − (SX)^{2}). e) Here n = 3, SX = 6, SY = 5, SXX = 14 and SXY = 12, so that we get

b = −1/3 and m = 1.

#13 Let P (t) = Ce^{at} be the US population (in millions) t years after 1960. Taking ln on both sides we get that ln P = ln C + at, so using the regression line method (either as outlined in the text or the formula from #12) on the data points (0,ln 180), (10,ln 206) and (20,ln 226) we get that b = ln C ≈ 5.20 and m = a ≈ 0.0114, so that ln P = 5.20+0.0114t or P = 181.277e^{0.0114t} is the desired approximation. Using this we find the approximation for 1990 is P (30) ≈ 255.026, which is fairly good.

#15 Using the formulas from #12 is a very good idea here, since otherwise it gets very painful. Let y(x) be the postage x years since 1920, so the data points are (0,0.02), (12,0.03), (38,0.04),... .

a) We get y = 0.0066x − 0.2135 and using this to find the postage in 2000, we get y(80) = .314, which doesn’t make much sense, since it is lower then what we have even now.

b) The reason why the answer is so bad is that the data doesn’t look linear. It looks linear after 1972 though.

c) Repeating the same process as in #13 we get that ln y = 0.0529x − 4.9329 and plugging in x = 80 we get that ln y = −0.7009 and so y = e^{−0.7009} ≈ 0.50. Not very nice, but more likely then part a).