Title MSTE 16 Solutions.pdf ✅

16 Differential Calculus: Maxima and Minima Solutions ▣ 1. Find the maximum value of y = −3x 2 + 6x − 2. [SOLUTION] y ′ = −6x + 6 At y ′ = 0, −6x + 6 = 0 → x = 1 At x = 1, y = −3(1) 2 + 6(1) − 2 ymaximum = 1 ▣ 2. A right circular cylinder having a volume of 1200 cm3 is to be lined with gold foil costing Php 1 per square centimeter on its curved surface and with silver foil costing Php 0.60 per square centimeter at the top and bottom. Find its height for minimum cost. [SOLUTION] V = πr 2h 1200 = πr 2h h = 1200 πr 2 Express the cost for lining, C = (1)(2πrh) + (0.6)(2πr 2 ) C = 2πrh + 1.2πr 2 C = 2πr ( 1200 πr 2 ) + 1.2πr 2 C = 2400 r + 1.2πr 2 C ′ = − 2400 r 2 + 2.4πr At C ′ = 0, r = 6.8278 cm. Therefore, the height is h = 1200 π(6.8278) 2 h = 8.1934 cm ▣ 3. A piece of wire with length 14.283 cm is cut into two pieces. One piece is formed into a square and the other into a circle. Find the length of the circular wire so that the sum of the areas of the square and the circle is a minimum. [SOLUTION]
Let the portions be x and y. x + y = 14.283 The side of the square is: P = 4s x = 4s → s = x 4 The radius of the circle is C = 2πr y = 2πr → r = y 2π For the total area, A = s 2 + πr 2 A = ( x 4 ) 2 + π ( y 2π ) 2 A = 1 16 x 2 + 1 4π y 2 • A. Differentiation A = 1 16 x 2 + 1 4π (14.283 − x) 2 A ′ = 1 8 x + −2 4π (14.283 − x) At A ′ = 0, then x ≈ 8, y ≈ 6.283 cm • B. Cauchy-Schwarz Inequality (a 2 + b 2 )(x 2 + y 2 ) ≥ (ax + by) 2 Equality holds when a x = b y [( x 4 ) 2 + π ( y 2π ) 2 ] (4 2 + (2π) 2 π ) ≥ (x + y) 2 A(16 + 4π) ≥ 14.2832 A ≥ 7.1414 The minimum area is 7.1414 sq cm. This occurs when the equality holds. x 4 4 = ( y 2π ) √π 2π √π x 16 = y 4π → x = 4 π y