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The most CREATIVE Math project EVER!!!
Christina Lawsonon 8 November 2012
Transcript of The most CREATIVE Math project EVER!!!
b) What would be the largest possible products if the sum of the numbers were k? Sixty meters of fencing are to be used to enclose a rectangular garden. a) Find the dimensions that will give the
b) What would be the maximum area if k feet of fencing were used? Forty feet of fencing are used to enclose three sides of a rectangular patio. The fourth side of the patio is a house and requires no fencing. a) Find the maximum area of the patio
b) What would be the maximum area if k feet of fencing were used? Work: x + y = 34 (sum of two numbers...) The largest possible product is 289.
The largest possible product (if the sum is k) is 1/2k*1/2k. WORK: 2x + 2y = 60 1 2x + 2y = 60 2 = x + y = 30 2 The dimensions that would give the maximum area would be 15 x 15.
If k feet of fencing were used k /4 would be the maximum area. WORK: 2x + 2y - x = 40 3 A grasshopper is h(t) = 8t -4t inches above the ground t seconds after it jumps. 2 a) How high does the grasshopper jump?
b) How long is the grasshopper in the air?
c) What relation does your answers to a and b have to the domain and range of h? 1. 2. 3. Create equations for the problem. xy = a Solve for y and plug in. y = -x+34 x(-x+34) = a Graph equation and find the maximum (vertex). -x +34x =a 2 -b/2a = -34/2(-1) = -34/-2 = 17 17 * 17 = 289 Create equation and simplify. Solve for y and plug in. y = -x + 30 xy = a x(-x + 30) = a -x + 30x = a 2 Graph equation and find maximum (vertex). -b/2a = -30/2(-1) = -30/-2 = 15 15 * 15 = 225 2 1 Simplify equation and solve for y. 2x + 2y = 40 + x 2 x + y = 20 + 1/2x y = -1/2x + 20 2 Graph and find the maximum. -b/2a = -20/2(-1/2) = -20/-1 = 20 20 x 20 = 400 ft. The maximum area of the patio if 400 ft.
If k feet of fencing were used the maximum area would be 1/2k * 1/2k WORK: 1 Graph equation and find maximum. f(t)= -4t + 8t 2 -b/2a = -8/2(-4) = -8/-8 = 1 2 Plug in x value of vertex to find y value. f(1) = -4(1) + 8(1) 2 v(1, 12) The grasshopper jumped 12 inches off the ground 1 second after it jumped. The domain represents time and the range represents the inches. Find the point on the graph of y = 2x + 10 that is close to the origin. How far is it? WORK: y = 2x + 10 1 Using the inverse slope, find the line perpendicular to y = 2x +10. y = -1/2x + 0 MAKE THE Y INTERCEPT 0 2 Graph and find the point of intersection between the two lines which is the closest point to the origin. Using trace, the intersection point is (-4, 2) 3 Use the distance formula to find how far it is away from the origin. (x - x ) + (y - y ) 2 2 1 1 (0-(-4)) + (0-2) 4-2 2 1.4 The point closest to the origin on y= 2x +10 is (-4,2) . Find the point on the line y = x + 1 closest to the point (1,0). How far is it? WORK: y = x + 1 1 Find the perpendicular line and graph. y = -x + 1 2 Find the point of intersection and use the distance formula. intersection point - (0,1) other point (1,0) The point (0,1) is the closet point on y = x + 1 to (1,0). A Motel has 20 rooms. If the manager charges $60 per room per night, all the rooms will be rented. For each $5 increase, one less room will be rented. a. How much rent should be charged to maximize revenue?
b. How many rooms are left empty to achieve this maximum? WORK: 1 Make revenue equations and look for a pattern. 20(60)= $1200
15(85)= $1275 or...you graph an equation and find the maximum of the parabloa which is still...$1280!
It looks like this: -5x +40x +1200 2 If a publisher charges $50 for a textbook, she will sell 8000 copies. For each $1 increase, she sells 100 fewer copies. a) How much should be charged to maximize revenue?
b) How many books will she sell at that rate? WORK: 50(8000)= $400,000
75(5500)= $412500 1 Come up with equations to calculate revenue and find max. or...come up with parabolic equation and graph to find max it looks like this: y = -100x + 3000x + 400,000 2 max= (15, 422500) To achieve maximum revenue, $80 should be charged per room.
4 rooms will be left empty. She should charge 65 dollars per book to maximize revenue.
6500 books would sell. I own 200 shares of Flybinite Enterprises that are worth $75 apiece. Each year I hold them, I get five free shares but the value of each share I have decreases by $2. a) How years should I hold these shares to maximize their total value?
b) Why did you need to consider the domain of your function to answer a) correctly? WORK: Create revenue equations and find maximum. 1 0 yr.- $75(200)= $15,000
1 yr.- $73(205)= $14,965
2 yr.- $71(210)= $14,910
3 yr.- $69(215)= $14,835 or...create a parabolic equation and graph the find the max. -10x -25x + 15,000 2 max= (-1.25, 15000) The shares should be held 0 yrs, to have maximum value.
The domain shows a gradual decrease in share value which helps determine max value. The revenue from the sale of x items is R(x)= 800x-2x and the cost to produce these x items is C(x)= 2x +1000 dollars. 2 2 a) How many items should be produced to minimize cost?
b) How many items should be produced to maximize revenue?
c) How many items should be produced to maximize profit?
d) If profit is to be maximized, what should be the selling price of each item? WORK:
1. Come up with a profit equation:
P(x)= (800x-2x )-(2x +1000) 2 2 p(x)= -4x + 800x + 1000 2 a. graph cost equation and find minimum. min= 0
b. Graph revenue equation and find max. max= 200
c. Graph profit equation and find max. max= 100 d. $39 PER ITEM.