Mat 160 - Assignment 3 Solutions - Summer 2012 - BSU - Jaimos F Skriletz 1 1. Limit Definition of te Derivative f( + ) f() Use te limit definition of te derivative, lim, to find te derivatives of te following functions: In te following I do te wole derivative in one step, but you are welcome to split it up into multiple steps. (a) f() = 3 5 2 f f( + ) f() [3( + ) 5( + ) 2 ] [3 5 2 ] 3 + 3 5 2 10 5 2 3 + 5 2 3 10 5 2 (3 10 5) (3 10 5) = 3 10 5(0) = 3 10 Tus te derivative is f () = 3 10 (b) f() = 1 2 Tus te derivative is f () = 2 (c) f() = m + b f f( + ) f() 1 (+) 2 1 2 2 (+) 2 2 (+) 2 2 2 2 2 2 ( + ) 2 1 (2 + ) 2 ( + ) 2 (2 + ) 2 ( + ) 2 (2 + 0) = 2 ( + 0) 2 = 2 4 = 2 f f( + ) f() [m( + ) + b] [m + b] m + m + b m b m m = m Tus te derivative is f () = m (Tis of course sould be te epected result since te derivative gives te slope and te slope of te line f() = m + b is m.)
Mat 160 - Assignment 3 Solutions - Summer 2012 - BSU - Jaimos F Skriletz 2 2. Power Rule Use te power rule and basic properties of te derivative to find te derivatives of te following functions: (a) P () = 12 5 7 6 + 52 (b) f() = 0.002 2.13 + 6.78 1.25 5 0.34 P () = 12(5 4 ) 7(3 2 ) 6 + 0 = 60 4 21 2 6 f () = 0.002(2.13 1.13 ) + 6.78(1.25 0.25 ) 5(0.34 0.66 ) = 0.00426 1.13 + 8.475 0.25 1.7 0.66 (c) f() = 6 12 = 6 1/2 12 1/2 f () = 3 1/2 + 6 3/2 = 3 + 6 3 (d) g() = 3 6 2 + 2 1 2 3. Quadratic Function = 3 2 62 2 + 2 2 1 2 = 6 + 2 1 2 g () = 1 2 2 + 2 3 = 1 2 2 + 2 = 3 2 + 2 Consider te quadratic function Q() = a 2 + b + c were a 0. (a) Te derivative is Q () = 2a + b (b) Te function as an orizontal tangent line if Q () = 2a + b = 0 2a = b = b 2a Tus tis quadratic as a orizontal tangent line (wic is te verte) at te point = b 2a.
Mat 160 - Assignment 3 Solutions - Summer 2012 - BSU - Jaimos F Skriletz 3 4. Tangent Lines Consider te function f() = 3 2 9 + 11. (a) Te derivative to te function is f () = 3 2 6 9 At te point = 1 te tangent line to f() will pass toug te point And te slope of te tangent line is Tus te equation of te tangent line at = 1 is (b) Te tangent line to f() is orizontal wen (1, f(1)) = (1, 0) m = f (1) = 12 y y 1 = m( 1 ) y 0 = 12( 1) y = 12 + 12 f () = 3 2 6 9 = 0 3( 2 2 3) = 0 3( + 1)( 3) = 0 = 1 or = 3 Tus te function as two points wic ave orizontal tangent lines, tose are ( 1, f( 1)) = ( 1, 16) and (3, f(3)) = (3, 16) (c) Te grap of f() wic sows te ow te info you found in te above problem relates to te grapical Properties of te function.
Mat 160 - Assignment 3 Solutions - Summer 2012 - BSU - Jaimos F Skriletz 4 5. Marginal Economics A company is planning to manufacture and market a four-slice toaster. For tis toaster, researc department s estimates are a weekly demand of 300 toasters at a price of $25 per toaster and a weekly demand of 400 toasters at a price of $20 per toaster. Te financial department s estimates are a fied weekly costs of $5,000 and a variable cost of $5 per toaster. Let p be te price per toaster and be te number sold eac week (weekly demand). (a) Assuming te relation sip is linear, we ave a line tat passes toug te two points ( 1, p 1 ) = (300, 25) and ( 2, p 1 ) = (400, 20) Tus te slope of te line is Tus te equation of te line is m = p 20 25 = 400 300 = 5 100 = 1 20 = 0.05 p p 1 = m( 1 ) p 25 = 0.05( 300) p 25 = 0.05 + 15 p = 0.05 + 40 Now we are only interested in te part of tis line tat is in te first quadrant. Tus we want to ensure tat bot 0 and p 0. Te second inequality gives Tus te domain is 0 800 p = 0.05 + 40 0 0.05 40 40 0.05 = 800 (b) Te revenue function is R() = p = ( 0.05 + 40) = 0.05 2 + 40 Te domain of tis function is te same as te demand, so 0 800. (c) Assuming te cost function is linear wit fied costs of $5,000 and a variable cost of $5 per toaster gives (d) Te break-even points occur wen C() = 5 + 5000 R() = C() 0.05 2 + 40 = 5 + 5000 0.05 2 + 35 5000 = 0 2 700 + 100000 = 0 ( 200)( 500) = 0 = 200 or = 500 Tus te two break even points occur at = 200 and = 500 as seen in te following grap.
Mat 160 - Assignment 3 Solutions - Summer 2012 - BSU - Jaimos F Skriletz 5 (e) Te profit function is (f) Te average profit is P () = R() C() = [ 0.05 2 + 40] [5 + 5000] = 0.05 2 + 35 5000 P () = P () If = 325 ten te average profit is P (325) = $3.37 per unit. If = 425 ten te average profit is P (425) = $1.99 per unit. (g) Te marginal profit is If = 325 ten te marginal profit is P (325) = $2.5. If = 425 ten te marginal profit is P (425) = $7.5. = 0.052 + 35 5000 P () = 0.1 + 35 () At a production level of = 325 units te average profit is $3.37 per unit, wile te profit would increase at an instantaneous rate of $2.50 per unit if you were to increase te production level. In contrast at a production level of = 425 units te average profit is $1.99 per unit, but te profit would decrease at an instantaneous rate of $7.50 per unit if you were to increase your production level.