Math 1151, Calculus ILecture XLIReview for the finalMath 1151, Calculus I Lecture XLIReview for the final 1 / 15The final exam will be given from 8:00 - 9:45pm on Tuesday December 11.Section 6373 (The morning section) will be in EL 1008.Section 9790 (The afternoon section) will be in SH 100.The Make up exam will be 8 - 9:45am Wednesday December 12 at SmithLab 1153.Math 1151, Calculus I Lecture XLIReview for the final 2 / 15substitutionComputeRex2xdxMake the substitution u = x2. so that du = 2xdx and xdx =12du.Rex2xdx =Reu12du =12eu+ C =12ex2+ CMath 1151, Calculus I Lecture XLIReview for the final 3 / 15substitutionComputeRex2xdxMake the substitution u = x2. so that du = 2xdx and xdx =12du.Rex2xdx =Reu12du =12eu+ C =12ex2+ CMath 1151, Calculus I Lecture XLIReview for the final 3 / 15substitutionComputeRex2xdxMake the substitution u = x2. so that du = 2xdx and xdx =12du.Rex2xdx =Reu12du=12eu+ C =12ex2+ CMath 1151, Calculus I Lecture XLIReview for the final 3 / 15substitutionComputeRex2xdxMake the substitution u = x2. so that du = 2xdx and xdx =12du.Rex2xdx =Reu12du =12eu+ C=12ex2+ CMath 1151, Calculus I Lecture XLIReview for the final 3 / 15substitutionComputeRex2xdxMake the substitution u = x2. so that du = 2xdx and xdx =12du.Rex2xdx =Reu12du =12eu+ C =12ex2+ CMath 1151, Calculus I Lecture XLIReview for the final 3 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 =7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15related ratesSuppose that the height of a rectangle changes at a rate of −3 and thatthe base changes at a rate of 4.At what rate does the area change when the height and width are both 7?Formula A = b · h.Differentiate:A0= b ·h0+ b0· hWe are given that h0= −3, b0= 4 and that at the moment of interestb = h = 7.Substitute in these know values:A0= 7 ·(−3) + 4 · 7 = 7Math 1151, Calculus I Lecture XLIReview for the final 4 / 15optimizationWe need to fence off a rectangular region. The fence we must use for theeast-west fence is $20 per foot. The north-south fencing is $15 per foot. Ifwe need to fence off 100 square feet, what is the cheapest option?We need to minimize cost = 2 ·20 · (height) + 2 · 15 ·(width)subject to the constraint that Area = (height) · (width) = 100.Minimize 40h + 30w with w · h = 100 w ≥ 0 and h ≥ 0.(1) Solve for w. w =100h. Minimize f (h) := 40h +3000h(2) h ≥ 0 and100h≥ 0 combine to give only h > 0 (we need to divide byh). NOT A COMPACT DOMAIN.(2) If there is a global min it will be a critical point: f0(h) = 40 −3000h2.40 −3000h2= 0, 40 · h2= 3000, h =r3004= 5√3 ∼ 8.66Only one critical point. If it is a local min then it is a global min:Math 1151, Calculus I Lecture XLIReview for the final 5 / 15optimizationWe need to fence off a rectangular region. The fence we must use for theeast-west fence is $20 per foot. The north-south fencing is $15 per foot. Ifwe need to fence off 100 square feet, what is the cheapest option?We need to minimize cost = 2 ·20 · (height) + 2 · 15 ·(width)subject to the constraint that Area = (height) · (width) = 100.Minimize 40h + 30w with w · h = 100 w ≥ 0 and h ≥ 0.(1) Solve for w. w =100h. Minimize f (h) := 40h +3000h(2) h ≥ 0 and100h≥ 0 combine to give only h > 0 (we need to divide byh). NOT A COMPACT DOMAIN.(2) If there is a global min it will be a critical point: f0(h) = 40 −3000h2.40 −3000h2= 0, 40 · h2= 3000, h =r3004= 5√3 ∼ 8.66Only one critical point. If it is a local min then it is a global min:Math 1151, Calculus I Lecture XLIReview for the final 5 / 15optimizationWe need to fence off a rectangular region. The fence we must use for theeast-west fence is $20 per foot. The north-south fencing is $15 per foot. Ifwe need to fence off 100 square feet, what is the cheapest option?We need to minimize cost = 2 ·20 · (height) + 2 · …