1. A bullet is shot upwards with an initial velocity of 1000 ft/sec from a point 20 ft above groand its height above the ground at time t is given by h(t)=-16t^2 + 1000t + 20. How high will the bullet go and how long will it take the bullet to reach the highest point?
2. A transit authority for a major metropolitan area has determined an approximate demand function which express daily ridership s a function of the fare charged (p). the demand function is D(p)=q=15000-120p. determine how many riders per day are expected under the fare that maximizes the revenue.
you are looking for the vertex of
h = -16t^2 + 1000t + 20
the t of the vertex is -1000/-32 = 31.25 seconds
for that value of t,
h = 15645 ft
thanks whats the second quetion
1. To find the height at which the bullet will go, we need to determine the maximum point of the given quadratic function h(t) = -16t^2 + 1000t + 20.
The maximum point of a quadratic function in the form ax^2 + bx + c is given by the formula:
t_max = -b / (2a)
In this case, a = -16 and b = 1000, so we can substitute those values into the formula:
t_max = -1000 / (2*(-16)) = -1000 / (-32) = 31.25 seconds
To find the height reached at this time, substitute the value of t_max into the height function:
h_max = -16(31.25)^2 + 1000(31.25) + 20
= -16 * 976.5625 + 31250 + 20
= -15625 + 31250 + 20
= 15645 feet
Therefore, the bullet will reach a maximum height of 15645 feet.
To find the time it takes for the bullet to reach the highest point, we can simply use the value of t_max, which is 31.25 seconds.
Hence, it will take the bullet 31.25 seconds to reach the highest point.
2. To find the fare that maximizes revenue, we need to determine the price for which the demand function D(p) = 15000 - 120p is maximized.
The formula for revenue is given by:
Revenue = Price (p) * Quantity (Q)
Since the quantity (Q) is represented by the demand function D(p), we substitute D(p) into the revenue formula:
Revenue = p * D(p)
= p * (15000 - 120p)
To maximize revenue, we need to find the value of p that maximizes this function. One way to do this is to take the derivative of the revenue function with respect to p and set it equal to zero:
d(Revenue) / dp = 0
Taking the derivative of p * (15000 - 120p) with respect to p gives:
15000 - 240p = 0
Solving for p:
-240p = -15000
p = 15000 / 240
p ≈ 62.5
Therefore, the fare that maximizes revenue is approximately $62.50.
To find the number of riders per day expected under this fare, we substitute the value of p into the demand function D(p):
Q = D(p)
= 15000 - 120p
= 15000 - 120(62.5)
= 15000 - 7500
= 7500
Hence, under the fare that maximizes revenue, there are expected to be 7500 riders per day.
1. To find out how high the bullet will go and how long it will take to reach the highest point, we need to find the vertex of the parabolic function h(t) = -16t^2 + 1000t + 20.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula x = -b / (2a) and y = f(x), where f(x) is the corresponding y-value.
In this case, we can see that a = -16, b = 1000, and c = 20. Plugging these values into the formula, we can calculate the time it takes for the bullet to reach its highest point:
t = -b / (2a)
t = -1000 / (2 * -16)
t = 31.25 seconds
Now, to find the height of the bullet at its highest point, we substitute the value of t into the equation h(t):
h(t) = -16t^2 + 1000t + 20
h(31.25) = -16(31.25)^2 + 1000(31.25) + 20
h(31.25) = -16(976.56) + 31250 + 20
h(31.25) = -15625 + 31250 + 20
h(31.25) = 15645 ft
Therefore, the bullet will reach a maximum height of 15645 ft, and it will take 31.25 seconds to get there.
2. To determine the number of riders per day expected under the fare that maximizes revenue, we need to find the fare that maximizes revenue first. Revenue is maximized when demand is maximized.
The demand function is given as D(p) = q = 15000 - 120p, where p is the fare charged and q is the number of riders per day.
To find the fare that maximizes revenue, we need to find the value of p that maximizes the demand function. This can be achieved by taking the derivative of the demand function with respect to p and setting it equal to zero.
dD(p)/dp = -120
-120 = 0
Solving for p, we find that p = 0.
Since we cannot have a fare of $0, we need to investigate further. We can examine the slopes of the demand function to determine if the maximum occurs at the boundaries.
At p = 0 (the lower boundary), the slope of the demand function is -120. As the fare increases towards the upper boundary, the slope continuously decreases.
Therefore, the fare that maximizes revenue is at the upper boundary, beyond which the slope becomes negative. However, since an upper boundary is not provided, we cannot determine the exact fare that maximizes revenue.
To compute the number of riders per day expected under the fare that maximizes revenue, we need more information about the upper boundary of the fare.