# The total time required for the light to travel from

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Unformatted text preview: − y )/8, y = 8 − 4x/3 so A = x(8 − 4x/3) = 8x − 4x2 /3 for x in [0, 6]. dA/dx = 8 − 8x/3, dA/dx = 0 when x = 3. If x = 0, 3, 6 then A = 0, 12, 0 so the area is greatest when x = 3 in and (from y = 8 − 4x/3) y = 4 in. 10 8 x y 6 7. 8. Let x, y , and z be as shown in the ﬁgure and A the area of the rectangle, then A = xy and, by similar triangles, z/10 = y/6, z = 5y/3; also x/10 = (8 − z )/8 = (8 − 5y/3)/8 thus y = 24/5 − 12x/25 so A = x(24/5 − 12x/25) = 24x/5 − 12x2 /25 for x in [0, 10]. dA/dx = 24/5 − 24x/25, dA/dx = 0 when x = 5. If x = 0, 5, 10 then A = 0, 12, 0 so the area is greatest when x = 5 in. and y = 12/5 in. z y 10 8 x 6 A = (2x)y = 2xy where y = 16 − x2 so A =√ x − 2x3 for 0√ x ≤ 4; 32 ≤ dA/dx = 0 when x = 4/ 3. If x = 0, 4/ √3, 4 dA/dx = 32 − 6x2 ,√ then A = 0, 256/(3 3), 0 so the area is largest when x = 4/ 3 and y = 32/3. The dimensions of the rectangle with largest area are √ 8/ 3 by 32/3. y 16 y x x -4 9. 10. √ A = xy where x2 + y 2 = 202 = 400 so y = 400 − x2 and √ √ ≤ dA/dx = 2(200 √ x2 )/ 400 − x2 , − A = x 400 − x2 for 0√ x ≤ 20; √ dA/dx = 0 when x = 200 = 10 2. If x = 0, 10 2, 20 then √ A = √, 200, 0 so the area is maximum when x = 10 2 and 0 √ y = 400 − 200 = 10 2. Let x and y be the dimensions shown in the ﬁgure, then the area of the rectangle is A = xy . x2 1 But + y 2 = R2 , thus y = R2 − x2 /4 = 4R2 − x2 so 2 2 √ 1 A = x 4R2 − x2 for 0 ≤ x ≤ 2R. dA/dx = (2R2 − x2 )/ 4R2 − x2 , 2 √ √ dA/dx = 0 when x = 2R. If x = 0, 2R, 2R then A = 0, R2 , 0 so √ √ the greatest area occurs when x = 2R and y = 2R/2. 10 4 y x x R y x 2 Exercise Set 6.2 11. 184 Let x = length of each side that uses the \$1 per foot fencing, y = length of each side that uses the \$2 per foot fencing. The cost is C = (1)(2x) + (2)(2y ) = 2x + 4y , but A = xy = 3200 thus y = 3200/x so C = 2x + 12800/x for x > 0, dC/dx = 2 − 12800/x2 , dC/dx = 0 when x = 80, d2 C/dx2 > 0 so C is least when x = 80, y = 40. 12. A = xy where 2x + 2y = p so y = p/2 − x and A = px/2 − x2 for x in [0, p/2]; dA/dx = p/2 − 2x, dA/dx = 0 when x = p/4. If x = 0 or p/2 then A = 0, if x = p/4 then A = p2 /16 so the area is maximum when x = p/4 and y = p/2 − p/4 = p/4, which is a square. y x 13. Let x and y be the dimensions of a rectangle; the perimeter is p = 2x + 2y . But A = xy thus √ y = A/x so p = 2x + 2A/x for x > 0, dp/dx = 2 − 2A/x2 = √ x2 − A)/x2 ,√ 2( dp/dx = 0 when x = A, d2 p/dx2 = 4A/x3 > 0 if x > 0 so p is a minimum when x = A and y = A and thus the rectangle is a square. 14. With x, y , r, and s as shown in the ﬁgure, the sum of the enclosed x y x and s = because x is the areas is A = πr 2 + s2 where r = 2π 4 circumference of the circle and y is the perimeter of the square, thus r x2 y2 A= + . But x + y = 12, so y = 12 − x and 4π 16 (12 − x)2 π+4 2 3 x2 A= + = x − x + 9 for 0 ≤ x ≤ 12. 4π 16 16π 2 π+4 3 dA 12π 12π dA = x− , = 0 when x = . If x = 0, , 12 dx 8π 2 dx π+4 π+4 36 36 then A = 9, , so the sum of the enclosed areas is π+4 π (a) a maximum when x = 12 in. (when all of the wire is used for the circle) (b) 15. (a) 17. y cut s a minimum when x = 12π/(π + 4) in. dN = 250(20 − t)e−t/20 = 0 at t = 20, N (0) = 125000, N (20) ≈ 161788, and N (100) ≈ 128,369; dt the absolute maximum is N = 161788 at t = 20, the absolute minimum is N = 125000 at t = 0. (b) The absolute minimum of 16. 12 dN d2 N occurs when = 12.5(t − 40)e−t/20 = 0, t = 40. dt dt2 The area of the window is A = 2rh + πr2 /2, the perimeter is 1 p = 2r + 2h + πr thus h = [p − (2 + π )r] so 2 A = r[p − (2 + π )r] + πr2 /2 = pr − (2 + π/2)r2 for 0 ≤ r ≤ p/(2 + π ), dA/dr = p − (4 + π )r, dA/dr = 0 when r = p/(4 + π ) and d2 A/dr2 < 0, so A is maximum when r = p/(4 + π ). r h 2r V = x(12 − 2x)2 for 0 ≤ x ≤ 6; dV /dx = 12(x − 2)(x − 6), dV /dx = 0 when x = 2 for 0 < x < 6. If x = 0, 2, 6 then V = 0, 128, 0 so the volume is largest when x = 2 in. 12 x x x x 12 − 2 x 12 x x x x 12 − 2 x 185 Chapter 6 18. The dimensions of the box will be (k − 2x) by (k − 2x) by x so V = (k − 2x)2 x = 4x3 − 4kx2 + k 2 x for x in [0, k/2]. dV /dx = 12x2 − 8kx + k 2 = (6x − k )(2x − k ), dV /dx = 0 for x in (0, k/2) when x = k/6. If x = 0, k/6, k/2 then V = 0, 2k 3 /27, 0 so V is maximum when x = k/6. The squares should have dimensions k/6 by k/6. 19. Let x be the length of each side of a square, then V = x(3 − 2x)(8 − 2x) = 4x3 − 22x2 + 24x for 0 ≤ x ≤ 3/2; dV /dx = 12x2 − 44x + 24 = 4(3x − 2)(x − 3), dV /dx = 0 when x = 2/3 for 0 < x < 3/2. If x = 0, 2/3, 3/2 then V = 0, 200/27, 0 so the maximum volume is 200/27 ft3 . 20. Let x = length of each edge of base, y = height. The cost is C = (cost of top and bottom) + (cost of sides) = (2)(2x2 ) + (3)(4xy ) = 4x2 + 12xy , but V = x2 y = 2250 thus y = 2250/x2 so C = 4x2 + 27000/x for x > 0, dC/dx = 8x − 27000/x2...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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