I don't get either of these equations. Please help me out!
1. Music Mart sells an average of 150 cassette tapes daily at $10.00 a tape. If 15 sales are lost for each $0.50 increase in unit price, what price will maximize profits? What will be the resulting daily profit?
2. S. Piker served a volleyball from 1 meter off the ground with an initial upward velocity of 8 meters per second. Write an equation for the height of the ball "h" in meters after "t" seconds. What will be the height of the ball after .5 seconds? What is the maximum height of the ball? If no one touches the ball, when will it hit the floor?
Here's some help on #1. Please post one question per post. You will get quicker response that way.
1. To get the profit, you need to know the dealer's cost of each cassette as well as the revenue (the amount of money taken in).
Let N = number of cassettes sold
N = 150 - 15*[(P - 10)/0.50]
= 450 - 30 P
P is the price in dollars. If P = 15, none are sold.
Revenue = R = NP = 450 P - 30 P^2
For maximum revenue, dR/dP = 0
450 - 60 P = 0
That results in P = $7.50 for maximum revenue, but you'll probably be losing money at that price. What you need is an equation for PROFIT vs. number sold, not total revenue. For that, you need the wholesale cost per cassette.
1. Music Mart sells an average of 150 cassette tapes daily at $10.00 a tape. If 15 sales are lost for each $0.50 increase in unit price, what price will maximize profits? What will be the resulting daily profit?
A few punches on the calculator will quickly expose the answer.
Alternatively,
Profit, P = (150 - 15x)(10 + .5x)
Multiplying out yields 7.5x^1 75x -1500.
Solve for the unexpected result using the quadratic formula.
2. S. Piker served a volleyball from 1 meter off the ground with an initial upward velocity of 8 meters per second. Write an equation for the height of the ball "h" in meters after "t" seconds. What will be the height of the ball after .5 seconds? What is the maximum height of the ball? If no one touches the ball, when will it hit the floor?
Judicious use of the equations of motion will take you to the answers.
The relationships between initial velocity, final velocity, distance covered and time, in uniformlay accelerated motion are defined mathematically by the following three equations:
1--From a = (Vo - Vf)/t, the final velocity of a body under constant acceleration is given by Vf = Vo + at.
2--The second equation regarding accelerated motion defines the distance traveled by a body under uniform acceleration.The average velocity of a moving body during a time interval "t" is expressed by Vav = (Vo + Vf)/2. The distance traveled, "s", during this time interval "t" is the product of the average velocity and the duraion of the time interval or s = (Vo + Vf)t/2. Substituting Vf = Vo + at into this expression yields s = [Vo + (Vo + at)]t/2 or s = Vot + at^2/2.
3--The third equation of uniformly accelerated motion is derived from the first two by eliminating the time interval "t". Multiplying the two expressions results in as = (Vf - Vo)/2 x (Vo + Vf)t/2 or Vf^2 = Vo^2 + 2as.
In summary,
Vf = Vo + at
s = Vot + at^2/2
Vf^2 = Vo^2 + 2as
These same equations apply to rising and falling bodies with the exception that a is replaced by g, the acceleration due to gravity.
For rising bodies,
Vf = Vo - gt
s = Vot - gt^2/2
Vf^2 = Vo^2 - 2gs.
For falling bodies,
Vf = Vo + gt
s = Vot + gt^2/2
Vf^2 = Vo^2 + 2gs
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Sure, let's break down each equation step by step:
1. To find the price that will maximize profits, we need to understand the relationship between price, sales, and profit.
First, we know that Music Mart sells an average of 150 cassette tapes daily at $10.00 per tape. This means that their current revenue (R) is $10.00 multiplied by 150, which gives us $1500.00.
Next, we need to consider the impact of increasing the unit price by $0.50. For every $0.50 increase, 15 sales are lost. This tells us that the number of tapes sold (S) will decrease by 15 for each $0.50 increase in price.
In general, we can express the number of tapes sold (S) based on the unit price (P) using the equation:
S = 150 - (15 * (P - 10) / 0.50)
To find the price that maximizes profits, we need to find the unit price (P) that corresponds to the maximum profit. Profit (Pr) is calculated by subtracting the cost (C) from the revenue (R):
Pr = R - C
Since the cost is not given, we'll assume it remains constant.
Now, we have an equation for profit:
Pr = (P * S) - C
By substituting the equation for S, we can express profit (Pr) in terms of P:
Pr = (P * (150 - (15 * (P - 10) / 0.50))) - C
To find the price that maximizes profits, you can graph this equation or use calculus to find the critical points. Set the derivative of Pr with respect to P equal to zero and solve for P:
dPr/dP = 0
Once you find the price (P), you can plug it back into the equation for S to find the resulting daily profit.
2. The equation for the height of the ball (h) in meters after time (t) seconds can be determined using kinematic equations:
h = h0 + v0t + (1/2)at^2
In this case, the initial height (h0) is 1 meter and the initial upward velocity (v0) is 8 meters per second. Assuming no air resistance, we can consider the acceleration (a) to be the acceleration due to gravity, approximately -9.8 meters per second^2.
So, the equation for the height of the ball becomes:
h = 1 + 8t - (1/2)(9.8)t^2
To find the height of the ball after 0.5 seconds, substitute t = 0.5 into the equation:
h = 1 + 8(0.5) - (1/2)(9.8)(0.5)^2
Simplifying, you'll get the height after 0.5 seconds.
To find the maximum height of the ball, you need to find the vertex of the quadratic equation. The vertex occurs at t = -b / (2a), where a and b are the coefficients of the t^2 and t terms, respectively.
For our equation, t = -8 / (2 * -4.9) simplifies to t = 0.816.
Substitute t = 0.816 back into the equation for height to find the maximum height.
Finally, to calculate when the ball hits the floor, set h = 0 and solve the quadratic equation for t. The positive root will give you the time it takes for the ball to hit the floor.