A lidless box is to be made using 2 m^2 of cardboard. Find the dimensions of the

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A lidless box is to be made using 2 m^2 of cardboard. Find the dimensions of the box with the largest possible volume. A detailed answer would be helpful

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Take-Home #06 – MATH 2421 Solutions • Section 12.6 12.6.20 Compute the directional derivative of f(x, y) = x x-y at the point P(4, 1) in the direction of h-1, 2i. Soln: The directional derivative is given by Duf(P) = rf(P)·u where u is a unit vector in the direction of h-1, 2i. This means that u = (1/p5) h-1, 2i rf(x, y) = (x - y)(1) - x(1) (x - y)2 , (x - y)(0) - (x)(-1) (x - y)2 = -y (x - y)2 , x (x - y)2 = 1 (x - y)2 h-y, xi so the directional derivative is Duf(P) = rf(P) · u = 1 9p5 h-1, 4i · h-1, 2i = 1 9p5 (1 + 8) = 1 p5 12.6.22ab (a) Find the unit vector that gives the direction of steepest ascent and descent at P. (b) Find a vector that points in the direction of no change in the function at P. f(x, y) = 6x2 + 4xy - 3y2; P(6,-1) Soln: (a) For the steepest ascent we find the unit vector parallel to the gradient of f at P, and for the steepest descent we find the negative of that unit vector. rf(x, y) = h12x + 4y, 4x - 6xi =) rf(P) = h12(6) + 4(-1), 4(6) - 6(-1)i = h68, 30i Therefore, the direction of steepest ascent is u = 1 p5524 h68, 30i and the direction of steepest descent is -u = -1 p5524 h68, 30i (b) To find a vector that points in the direction of no change we simply need a vector that is perpendicular to the gradient. Any vector parallel to v = h-30, 68i would work. 12.6.40 Consider the paraboloid f(x, y) = 16 - x2/4 - y2/16 and the point P(2p3, 4) on the level curve f(x, y) = 12. Compute the slope of the line tangent to the level curve at P and verify that the tangent line is orthogonal to the gradient at that point. Soln: To find the slope of the level curve at P we simply need dy/dx = -Fx/Fy (by the implicit function theorem). Let F(x, y) = f(x, y) - 12 and consider the level curve F(x, y) = 0. dy dx = - -x/2 -y/8 = - 4x y = - 4x y Evaluated at the point P we get dy dx = - 8p3 4 = -2p3 The direction of the tangent is parallel to 1,-2p3 . Check: The gradient of f at this point is rf(2p3, 4) = -p3,-1/2 . This is parallel to 2p3, 1 which is perpendicular to the tangent direction. 12.6.58b Consider the function F(x, y, z) = exyz. (b) relate rF to rg where F = f g, f is a function of one variable, and g is a function of three variables. 1 Soln: From the hint on the HW sheet, f(u) = eu and g(x, y, z) = xyz. rF = hyzexyz, xzexyz, xyexyzi = exyz hyz, xz, xyi and rg = hyz, xz, xyi Since exyz is a scalar that can be factored out of rF we see that rF is parallel to rg . Also notice that rF(x, y, z) = f0(g(x, y, z))rg(x, y, z) • Section 12.7 12.7.17 Find an equation of the plane tangent to the surface f(x, y) = x2ex-y at the points (2, 2, 4) and (-1,-1, 1). Soln: To find the equation of a tangent plane we use the formula f(x, y) f(a, b) + fx(a, b)(x - a) + fy(a, b)(y - b). Hence, we need to find the partial derivaties. fx(x, y) = x2ex-y + 2xex-y = ex-y(x2 + 2x) fy(x, y) = -x2ex-y Next we plug in the point (2, 2, 4): fx(2, 2) = 1(4 + 4) = 8, and fy(2, 2) = -4. Lastly, f(2, 2) = 4. Therefore, the tangent plane is z = 4 + (8)(x - 2) - 4(y - 2) Next consider the point (-1,-1, 1): fx(-1,-1) = -1, fy(-1,-1) = -1, f(-1,-1) = 1 so the equation for the tangent plane is z = 1 - (x + 1) - (y + 1) 12.7.30 Use differentials to approximate the change in z for given changes in x and y. f(x, y) = ln(1 + x + y) when (x, y) changes from (0, 0) to (-0.1, 0.03). Soln: First we need to write the total differential: df = @f @x dx + @f @y dy = 1 1 + x + y dx + 1 1 + x + y dy In this context, dx = -0.1 and dy = 0.03 starting at the point (0, 0). This means that dz = 1 1 + 0 + 0 (-0.1) + 1 1 + 0 + 0 (0.03) = -0.1 + 0.03 = -0.07 12.7.32a The volume of a right circular cone with radius r and height h is V = r2h/3. (a) Approximate the change in volume of the when the radius changes from r = 6.5 to r = 6.6 and the height changes from h = 4.20 to h = 4.15. 2 Soln: (a) First find the total differential: dV = @V @r dr + @V @h dh = 2rh 3 dr + r2 3 dh. In this case, dr = 0.1 and dh = -0.05 starting at the point (r, h) = (6.5, 4.20). Therefore, dV = 2(6.5)(4.2) 3 (0.1) + (6.5)2 3 (-0.05) = 3 ?? 2(6.5)(4.2)(0.1) + (6.5)2(0.05) = 3 (5.46 - 2.1125) = 3 (3.3475) · 1.11583 3.504 12.7.40 The cost of a trip that is L miles long, driving a car that gets m miles per gallon, with gas costs $p/gal is C = Lp/m dollars. Suppose you plan a trip of L = 1500mi in a car that gets m = 32mi/gal, with gas costs of p = $3.80/gal. (a) Explain how the cost function is derived. (b) Compute the partial derivatives CL, Cm, and Cp. Explain the meaning of the signs of the derivaties in the context of the problem. (c) Estimate the change in the cost of the trip if L changes from L = 1500 to L = 1520, m changes from m = 32 to 31, and p changes from $3.80 to p = $3.85. (d) Is the total cost of the trip (with L = 1500mi, m = 32mi/gal, and p = $3.80) more sensitive to a 1% change in L, m, or p (assuming the other variables are fixed)? Explain. Soln: (a) The cost function is obtained by multiplying the distance by the cost per mile. (b) Partial derivatives: CL = p m , Cm = - Lp m2 , Cp = L m Signs: Since L, p, and m are all positive, C is an increasing function of L and p, and C is a decreasing function of m. (c) Use the total differential dC = CLdL + Cmdm + Cpdp = p m dL - Lp m2 dm + L m dp Here we know that dL = 20, dm = -1, and dp = $0.05. Plugging these into the total differential at the point (1500, 32, 3.80) gives dC = (3.8) (32) (20) - (1500)(3.8) (32)2 (-1) + (1500) (32) (0.05) $10.29 (d) First let us write the relative change dC/C. dC C = CL C dL + Cm C dm + Cp C dp = dL L - dm m + dp p Since each of these coefficients is 1 or -1 the cost is equally sensitive to changes in each of the three variables. In other words, a k% change in on variable while holding the other two fixed will yield a k% change in C. • Section 12.8 12.8.24 Find the critical points of the following functions. Use the second derivatives test to determine (if possible) whether each critical point corresponds to a local max, min, or saddle point. Confirm your results with a graphing utility. f(x, y) = x2 - x4/2 - y2 - xy Soln: To find the critical points we set the gradient vector equal to the zero vector. rf(x, y) = 2x - 2x3 - y, -2y - x 3 Setting this to zero gives the system of equations 2x - 2x3 - y = 0 -2y - x = 0 The second equation indicates that y = -x/2. Plugging this into the first equation gives 2x - 2x3 + x/2 = 0 =) 4x - 4x3 + x = 0 =) -4x3 + 5x = 0 =) x(4x2 - 5) = 0 Hence, x = 0, x = ±p5/2. Plugging this into the equation y = -x/2 gives y = 0, p5/4. Critical Points: (0, 0), p5 2 ,- p5 4 ! , - p5 2 , p5 4 ! Now we need to find the classification for each of the critical points. First find the second derivatives: fxx(x, y) = 2 - 6x2, fyy = -2, fxy = fyx = -1 This means that D(x, y) = fxxfyy - f2 xy = (2 - 6x2)(-2) - 1 = -4 + 12x2 - 1 = 12x2 - 5. Crit. Pt. fxx D Conclusion (0, 0) 2 < 0 Saddle Point (p5/2,-p5/4) < 0 > 0 local max (-p5/2,p5/4) < 0 > 0 local max 12.8.25 (same directions as #24). f(x, y) = x - y 1 + x2 + y2 Hint: @ @x x - y 1 + x2 + y2 = 2xy - x2 + y2 + 1 (x2 + y2 + 1)2 , @ @y -2xy - x2 + y2 - 1 (x2 + y2 + 1)2 Soln: Find the gradient and set it to the zero vector first rf(x, y) = (1 + x2 + y2)(1) - (x - y)(2x) (1 + x2 + y2)2 , (1 + x2 + y2)(-1) - (x - y)(2y) (1 + x2 + y2)2 = 1 (1 + x2 + y2)2 1 - x2 + y2 + 2xy, -1 - x2 + y2 - 2xy Now to find the critical points we must solve the simultaneous system of equations fx = 0 and fy = 0. (Note that there are no points where the partial derivatives are undefined). y2 + 2xy - x2 + 1 = 0 y2 - 2xy - x2 - 1 = 0 Adding these equations together gives y2 - x2 = 0 which is the same as y = ±x. Subtracting them gives 2xy + 1 = 0. Now plug y = ±x into 2xy + 1 - 0 to get the equation ±2x2 + 1 = 0 which implies that x = ± 1 p2 . Hence, the critical points are CriticalPoints : 1 p2 , - 1 p2 and - 1 p2 , 1 p2 Now we need all of the second partial dervatives. After some simplifying we get fxx(x, y) = 2 ?? x3 + y - 3x2y + y3 - 3x ?? 1 + y2 (1 + x2 + y2)3 fyy(x, y) = - 2 ?? x + x3 - 3y - 3x2y - 3xy2 + y3 (1 + x2 + y2)3 fxy(x, y) = 2 ?? x + x3 + 3x2y - 3xy2 - y ?? 1 + y2 (1 + x2 + y2)3 Now we need to classify each of the critical points using the second derivative test. 4 Crit. Pt. fxx D Conclusion (1/p2,-1/p2) < 0 > 0 local max (-1/p2, 1/p2) > 0 > 0 local min 12.8.30 A lidless box is to be made using 2m2 of cardboard. Find the dimensions of the box with the largest possible volume. Soln: Let x, y > 0 be the dimensions and let h 0 be the height of the box. This means that the box has total area A = 2xh + 2yh + xy = 2 =) h = 2 - xy 2x + 2y . The volume of the box is given by V = xyh. Plugging in the equation for h gives V = (xy)(2 - xy) 2x + 2y = 1 2 xy(2 - xy) x + y . We need to minimize this function over the region R = {(x, y) : x 0, y 0, and xy 2} We will find the critical points in this region and compare the values at the critical points to the values on the boundary of the region. Notice that if (x, y) is on the boundary of R then the volume is zero. Therefore for this problem we only need to find the critical point(s) inside the region. Vx = 0 Vy = 0 =) ( 1 2 (2-x2 -2xy)y2 (x+y)2 = 0 1 2 (2-y2 -2xy)x2 (x+y)2 = 0 Notice that if the denominator is 0 then (x, y) = (0, 0). This yields a minimum for the volume of the box and hence we are not interested in this critical point. Notice that if y = x then the system of equations will be satisfied. Plugging this in to the denominator of the Vx gives (-3x2 + 2)(x2) = 0 which implies that x = 0 and x = ± p 2/3 are the critical points. The only point in the interior of R is x = p 2/3. This means that the only critical point is (x, y) = ( p 2/3, p 2/3). The corresponding height for this point is h( p 2/3, p 2/3) = p6/6. Therefore, the largest box has volume x = r 2 3 , y = r 2 3 , h = p6 6 =) V = p6 9 5

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