University:
Montgomery College
Course:
MATH 150: Elementary Applied Calculus I
Academic year:

2019
Views:

69

Pages:

4
Author:

Edgar Rowe

Name Solution Solve the problem. 1) If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where (3fb) p(x) = 61 - x/24 How many candy bars must be sold to maximize revenue? R(x) = xl =6|x- x^ eta 2 4 R' = 6i - (2x)/24 =^ 61 = 2X 24 2X = 1464 X=732 thousand Candy Bans pts) 2) Find the number of units that must be produced and sold in order to yield the maximum profit, given the following equations for revenue and cost: R(x) = 40x - 0.5x ^ 2 P(x) = R - C = (4x - 0.5x ^ i) - (8x + 5) P(x)=4 x - 0.5x ^ 2 - 8x - 5 P(x)=- bullet 0.5x ^ eta + 32x - 5 P' * (x) = - X + 32 = bullet X=32 units (4Pts) 3) A rectangular field is to be enclosed on sides with a fence. Fencing costs $3 per foot for two opposite sides, and $2 per foot for the other two sides. overleftrightarrow Find the dimensions of the field of area 850 ft2 that would be the cheapest to enclose. y X y X=35.71 feet y=23.80 feet X Xy =85 Rightarrow y= 8<= x Cost 3y+3y+2x+2x 6 st = 6y + 4x =6( 85o x )+ ^ 4 x Cost = 5100x + 4x 4X (Gst)^ prime = - 5iso * x ^ - 2 +4 Rightarrow -5i oo x^ 2 +^ 4 =0 A-1 5100 x2=+ 4x = 5100 2 x²=1275 X = 35.71 40,000 people per game. For every increase of $1, it loses 5,000 people. What price per ticket should be charged in order to maximize revenue? ( 4000 ,r10) (35000 m= |l-10 3zeta*oo -4 = i -5 =-o. 2 Y-10=-0,0002 (X-40000) y = -0,0002X+18 P = 0, 2x + 18 Revenue = XP = -0,0002x² +18x R ^ I =-0.0004 x+i8= bullet x =H5000 price = -0,0002 (45000) +18=9 3P) 5) A carpenter is building a rectangular room with a fixed perimeter of 440 ft. What are the dimensions of the largest room that can be built? What is its area? 2y + 2x =44 Rightarrow x+y=22 Rightarrow y=22 -x Area = xy = X(220-X) = 220x-x² Aren ^ prime = 22 ^ s - 2x = bullet X = 110 feet y=220-110 = 110 feet Area = (110) (110) = 12100 feet 2) Pts) 6) A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 53f * t ^ 3 What dimensions yield the minimum surface area? Round to the nearest tenth, if necessary. x^ 2 y=53 Rightarrow y = 53/(x ^ 2) beta len =x^ 2 +4xy=x^ 2 +4x( 53 x^ 2 )=x^ 2 +2i2 x^ -i (Arec)= 2x - 212x ^ - 2 = 0 2x 1 = 212 x^ 2 Rightarrow 2x ^ 3 = 212 4.7 feel by 4.7 feet by 2 ff feet/A-2 y = 2.4ft Solve the problem. (3pts) 1) If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in a certain city, where P(x) = 125 - x/18 R=X(125- x 18 )=12.5x- x^ 2 18 Rightarrow R' * (x) = 125 - 2/18 * X = 0 125 = X 9 ⇒ x = 1125 thousand Candy Bas (3pts) 2) Find the number of units that must be produced and sold in order to yield the maximum profit, given the following equations for revenue and cost: R(x) = 50x - 0.5x ^ 2 C(x) = 9x + 6 P = R - C = (50x - 0.5x ^ 2) - (9x + b) = 44x - 0.5x ^ 2 - 6 p' = 41 - ix = x = 41 units (4Pts) 3) A rectangular field is to be enclosed on four sides with a fence. Fencing costs $6 per foot for two opposite sides, and $3 per cheapest to foot for the other two sides. Find the dimensions of the field of area 690f * t ^ 2 that would be the enclose. xy =b9 o Rightarrow y= 69 o x y K y X 6s ^ t = 6y + 6y + 3x + 3x 66t = 12y + 6x Cost = 12(690)+6x = 8280x+6x \mathfrak{c}*s * t' = (- 828o)/(x ^ 2) + b =^ 6x ^ 2 = 828o x ^ 2 = 1380 X=37.15 feet y= 69 0 x Rightarrow sqrt[9] =18.57fegt) B-1