Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan. Solution to Section 4.5

Size: px

Start display at page:

Download "Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan. Solution to Section 4.5"

Stuart Townsend
5 years ago
Views:

1 Arkansas Tech University MATH 914: Calculus I Dr Marcel B Finan Solution to Section 45 1 (a) y y The two numbers with maimum product are 11 and 1 (b) Let f() = (3 ) where 0 3 f () = 3 = 0 = = 3 f has a local (and global) maimum at (115, 135) Also, f(0) = f(3) = 0 Hence, the two numbers are = y = Let the two numbers be and 100 We want the minimum of the function S() = + 100, > 0 We have f () = = 0 = = 10 Note that 1

2 f changes sign from negative to positive as crosses 10 from left to right Hence, S has a local minimum at = 10 Moreover, lim 0 + S() = and lim S() = Hence, f has a global minimum at = 10 The two numbers are = 10 and y = 10 5 Let D() denote the vertical distance Then D() = +, 1 Now, D () = 1 = 0 = = 05 Also, D (05) = < 0 so that D has a local maimum at = 05 On the other hand, D( 1) = 0 and D() = 0 so that D has a global maimum at = 05 and D(05) = 5 7 Let L denote the length of the rectangle and W its width We know that L + W = 100 so that X = 50 L The area of the rectangle is A(L) = L(50 L), 0 L 50 We have da dl = 50 L = 0 = L = 5 d Also, A (5) = < 0 so that A has a global maimum at L = 5 Hence, dl the dimensions of the rectangle are 5 m 5 m 11 Let be the length of the side of the base and y the height of the bo We have V = y and S = + 4y = 100 = y = Hence, V () = = V () = = 0 = = 0 Since V (0) = 30 < 0, V has a global maimum when = 0 Hence, y = 10 The largest possible volume is V = (0) (10) = 4000 cm 13 (a) Let and y be the sides of the rectangle We have A = y, where A is constant We want to minimize P = + y Since y = A, we have y = A Thus we have P () = + A Setting P () = 0, we have 0 = A Thus = ± A Since side lengths of rectangles are positive, we have = A Note that for 0 < < A, P () < 0 and for > A, P () > 0 Thus, P has a minimum at = A In this case, y = A and the rectangle with smallest perimeter and area A is a square (b) Let and y be the sides of the rectangle Then P = + y where P is a constant Hence, y = P We want to maimize A() = y = P We have A () = P = 0 = = P 4 Since, A () = < 0, A has a maimum at = P 4 In this case, y = P 4 and we have that of all rectangles with a given perimeter, the one with greatest area is a square 15 Let (, y) be a point on the line Then (, + 3) and the distance from this point to the origin is D() = + ( + 3) We want to minimize this function We have D 5+6) () = = 0 = = 6 +(+3) 5 Observe that the shortest distance to the origin must be a point on the line between the

3 intercept and the y intercept (see figure below) Hence, 3 0 Since D( 3/) = 3/, D(0) = 3, and D( 6/5) =, D has a global minimum at = 6/5 In this case, y = ( 6 5) + 3 = Let (, y) be a point on the ellipse The distance from this point to the point (1, 0) is D() = ( 1) + y = ( 1) = 5 3, 1 1 We have D () = 1 3 = = Since D () = 16(5 3 ) 3 < 0, D has a maimum at = 1 3 Since D( 1) = 4, D(1) = 0, and D( 1/3) 53, D has a global maimum at = 1 3 In this case, y = ± Let be the distance between the center of the circle and the base of the isosceles triangle Then the base of the triangle has length r and the height of the triangle is r + The area of the rectangle is A() = (r + ) r, r r We have A () = r r r = 0 = r r = 0 = = r Since A ( r ) = 9 ( ) < 0, A has a maimum at = r Since, A( r) = A(r) = 0, and A( r ) = r Hence, A has a global maimum at = r Hence, the base of the triangle is r 3 The other sides of the triangle have length 3r Hence, the isosceless triangle is an equilateral triangle 5 : The basic shape of the window is given by the following figure 3

4 To maimize to greatest amount of light admitted, we must maimize the area of the window Let be the width of the rectangle and y be the length Note that the radius r of the semicircle is half the width of the rectangle, that is, r = To find the objective area equation, we must find the area of the rectangle and the area of the semicircle The area of the rectangle is y and the area of the semi-circle is π 8 Thus, the total area is A = y + π 8 On the other hand, the perimeter around the window is 30 ft = + y + π = y = 15 π 4 Hence, the total area is A() = 15 π 8 Now, A () = 15 π 60 4 = 0 = = 4+π Since A () = 1 π 4 < 0, A has a global maimum at = π Hence, y = 15 4+π 15π π+4 = 30 4+π 7 Let the amount cut for the square be then the amount left for the equilateral triangle is 10 The area for the square will then be 16 and the area of the equilateral triangle will be 1 3(10 ) (10 ) 6 3 = 3(10 ) 36 So the total area enclosed is A() = 3(10 ) We have A () = 8 3(10 ) 18 = 0 = 435 Since A is a parabola that opens up, A has a global minimum at = 435 m On the other hand, A(0) 481 and A(10) 65 and A(435) 7 (a) The maimum area occurs when = 10 m and the entire length is used for the square 4

5 (b) The minimum area occurs when 435 m Let I denote the illumination and the distance from the object to the weaker source, where 0 < < 10 Then I() = k (10 ) + 3k We have I () = k 6k = 0 = = (10 ) Since, 3 I () = 6k + (10 ) 4 18k > 0, I has a minimum at = Also, lim 3 0 I() = lim 10 I() = so that I has a global minimum at = Let m be the slope of the line Clearly, m < 0 The equation of the line is y 5 = m( 3) The intercepts of the line are (0, 5 3m) and (3 5 m, 0) Therefore, the area of the triangle cut out by the line is A(m) = 1 (30 9m 5 m ) We have A (m) = 1 5 ( 9 + ) = 0 = m = 5 m 3 Since A (m) < 0, A has a maimum at m = 5 3 Also, lim m A(m) = lim m 0 A(m) = and so A has a global maimum at m = 5 3 The equation of the line is y 5 = 5 3 ( 3) or y = Suppose that the line is tangent to y = 3 at the point (a, 3 a ) Then the slope of the tangent line at that point is 3 and the equation of the a 5

6 tangent line is y 3 a = 3 ( a) The intercepts ar (0, 6 a a ) and (a, 0) Hence, the length of the line segment connecting the intercepts is L(a) = ( 4a + 36 ) 1 a We have L (a) = ( 4a 36 ) ( a 4a + 36 = 0 = a = 3 Since A (a) = ( ) ( a 5 a 4 a 4a + 36 ) 1 > 0 and lim a a 0 L(a) = lim a L(a) =, L has an absolute minimum at a = 3 The shortest length is L( 3) = 6 a ) 1 43 (a) Finding the derivative, we obtain c () = C () C() If c() has a minimum then we must have c () = 0 and this implies C () = c() That is, the average cost is equal to the marginal cost = C () c() (b) (i) C(1000) = , c(1000) = = 3449, and C (1000) = (ii) We must have C () = c() = = = = 400 (iii) The minimum average cost is c(400) = C(400) 400 = 30 per unit (a) Since p() is linear, we have p() = m+b ALso, m = = Hence, p() = But p(10) = 7000 = 7000 = b = b = 19 Hence, p() = (b) R() = p() = , 0 55, 000 We have R () = = 0 = = 8500 Since R() is a parabola that open down then R has a global maimum at = 8500 In this case the ticket price is p(8500) = = $950 6

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Page ( ) Water seeps out of a conical filter at the constant rate of 5 cc / sec. When the height of water level in the cone is 5 cm, find the rate at which the height