2.
f ( x) = x − 45 x + 600 x + 20 3 2a) Find the local extrema and identify them as either a local maximum or a local minimum.b) Find the coordinates of the absolute maximum and absolute minimum of the function in the interval [ 0,30]
3.
First,f ( x) = 3 x − 90 x + 600 2 differentiate. 0 = 3 x − 90 x + 600 2 Set f’(x) to 0 0 = x − 30 x + 200 2 Solve for x 0 = ( x − 10)( x − 20) Factor Hence, we have local extrema at x = 10 and x = 20
4.
To identify them as either maxima or minima, we can usethe derivative - f (9) = 3(9) 2 − 90(9) + 600 = 33 f (11) = 3(11) − 90(11) + 600 = −27 3
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To identify them as either maxima or minima, we can usethe derivative - f (19) = 3(19) 2 − 90(19) + 600 = −27 f (21) = 3(21) − 90(21) + 600 = 33 3
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b) Check the endpoints, x = 0 and x = 30 f (0) = (0) 3 − 45(0) 2 + 600(0) + 20 = 20 f (30) = (30) − 45(30) + 600(30) + 20 = 4520 3 2
7.
A rectangular pen is to be fenced in using two types of ncing. Two opposite sides will use heavy duty fencing at $3/ft hile the remaining two sides will use standard fencing at $1/ft.What are the dimensions of the rectangular plot of greatest area at can be fenced in at a total cost of $3600? A = xy 2 ( 3x ) + 2 ( 1y ) = 3600 3x + y = 1800 1y y = 1800-3x A = x ( 1800 − 3x ) 3x A = 1800x − 3x 2 A = 1800 − 6x A " = −6 0 = 1800 − 6x Therefore max 300 = x 900 = y The dimensions of a rectangular plot of greatest area are 300 x 900
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4. An open-top box with a square bottom and rectangular sidesis to have a volume of 256 cubic inches. Find the dimensionsthat require the minimum amount of material. xS = x + 4xy 2 256 → S = x + 4x 2 2 yV = x y = 256 2 x x 1024 y = 256/x² S = x2 + x 1024 2048 S = 2x − 2 S" = 2 + 3 > 0 x x 1024 therefore a min 0 = 2x − 2 x x=8 →y=4 8x8x4
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