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Ribbon Gutter

February 14th, 2010 |

Optimization Calculus Problems … Please Help Me!!?

Hello, I have three problems that have been a big problem for me. Can anyone help? If you do not show too much to ask as many steps as possible. Thank you very much:) 1) The cost of a particular model of car is $ 15000. When the dealer sells every car for $ 25000, sold 24 cars a month. For every $ 600 reduction in the selling price of the dealer sells more cars every 2 months. Determine the selling price of a car The maximum monthly benefit. 2) A piece of Christmas ribbon 100 cm long is cut into two pieces. A piece that forms a square and a circle forms. How should Martine Nigel and cut the ribbon for the total area enclosed is a maximum? 3) An isosceles traezoid drainage channel should be done, believing that the angles at A and B, on the cross for each of 120 degrees. If the 5m long sheet metal that has bebent to form covered open the stream has a width of 60cm, and then find the dimensions so that the cross-sectional area shall be a maximum. Thank you!

For a retail price of $ x, the number of cars sold increases (25000 – X) / 600 ie, number = 24 + (25000 – x) / 600 ……………… = (14400 + 25000 – x) / 600 ……………… = (39400 – x) / 600 P = Profit (number sales) * (x – 15000) ………… = (X – 15000) (39400 – x) / 600 P '= [39400 - x + 1 (x - 15000)] / 600 using the product rule … . = [54,400 - 2x] / 600 P '= 0 when x = 27,200. P "= -1/300, therefore the stationary value is a maximum turning point. However, the number of cars sold must be an integer, which is a small problem! The solution tells us that sells 20.3333 … cars for a total profit of $ 248,066.67. I suppose in practice figures would be in a selling price of $ 26,800 and $ 27,400, with sales of 21 cars and 20 cars, and found that $ 27,400 gives the most benefit. 2) That there is x cm for the square, leaving (100 – x) for the circle. Then, the square has side x / 4, zone x ² / 16. The circle has a radius (100-x) / 2π, π area (100-x) ² / 4π ² = (100-x) ² / 4π differentiate a '= 2x/16 + -2 (100-x) / 4π .. = X / 8 – (100 – x) / 2π solve a '= 0: I think it gives x = 400 / (π +4) However, note that the expression for the area is a second-degree positive coefficient of x ², which means that it is concave upward so that its apex is a minimum turning point. To confirm this finding and the second derivative is positive since, and therefore the stationary value is a minimum. The maximum area (795.77 square cm, approximately) is obtained when the entire tape is used to make a circle — there is no space at all! If all the tape is used for the square, the area is 625 square inches, and minimum area, using the value of x is obtained (about 56 cm) is 350 cm square. NOTE answers first made an error in the line A-625 = pi r ^ 2 + r ^ 2 PI / 4 + PI r ^ 2 Second quarter should be -25 πr 3) The length of the sloping edges be x cm. Then the length of the base is 60 – 2x cm. The height is x sin (60 °) x = √ 3 / 2 and the distance at the top is open (60 – 2x) + 2 * x cos (60 º) = 60 – x so the area (using (h / 2) (A + B)) is (x √ 3 / 4) (120 – 3x) A = (√ 3 / 4) (120x – 3x ²) I guess if this is a calculation exercise is necessary to differentiate and solve a '= 0 etc, but from knowledge of the parables, we can see that this will have zeros at 0, 40 and maximum value at x = 20 so that the metal has to be folded into three equal parts, each of width 20 cm.

Gutter Rats(UK) performing Shread to ribbons@Inferno Palooza v,mini fest.Bournemouth 11/04/09