Can someone help me with these questions?
1.What are the maximum and minimum values for Y=28(1.21)^x on the interval 0<x<12?
a) 275.8 and 0
b) 275.8 and 28
c) 249.7 and 28
d) 188.4 and 28
2. For what value of x does the maximum or minimum of f(x)= x^2 +8x + 12 occur? Is the point a maximum or minimum?
a) -2; maximum
b) -6; maximum
c) -4; minimum
d) 4; minimum
3. The monthly revenue for printer ink for a large office supply store is R(x) = -x^2 + 12x + 13, where x is the price of an ink cartidge in dollars and R(x) is in hundreds of dollars. What is the maximum monthly ink revenue?
a)$49
b)$3600
c)$4900
d)%12, 100
Thank you~ help needed pls T.T
since n^0 = 1 for any nonzero n,
28(1.21^x) on 0 < x < 12 goes from
28(1) to 28(9.849) = 28 to 275.77
so (B)
______________________
since the x^2 term has a positive coefficient, the parabola open upward, meaning it has a minimum
This minimum occurs when x = -b/2a = -8/2 = -4
so, (C)
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max occurs when x = -b/2a = -12/2 = 6
R(6) = -36 + 72 + 13 = 49
so (C)
is it $49 or 4900? you said R(6) = -36 + 72 + 13 = 49
Sure, let's go through each question step-by-step:
1. To find the maximum and minimum values of the function Y=28(1.21)^x on the interval 0<x<12, we can start by finding the derivative of the function with respect to x, which we'll call Y'(x).
Step 1: Find the derivative of Y(x):
Y'(x) = 28 * ln(1.21) * (1.21)^x
Step 2: Set Y'(x) to zero and solve for x to find critical points:
28 * ln(1.21) * (1.21)^x = 0
Since ln(1.21) is not equal to zero, we can divide both sides of the equation by 28 * ln(1.21):
(1.21)^x = 0
However, (1.21)^x can never be equal to zero, so there are no critical points.
Step 3: Determine the endpoints of the interval:
Since 0 < x < 12, the function is defined at both ends of the interval. So we need to evaluate the function at x=0 and x=12 to find the maximum and minimum values.
Y(0) = 28 * (1.21)^0 = 28
Y(12) = 28 * (1.21)^12 ≈ 275.8
So, the maximum value is approximately 275.8 and the minimum value is 28.
Therefore, the answer is option (b) 275.8 and 28.
2. To find the value of x at which the maximum or minimum of the function f(x) = x^2 + 8x + 12 occurs, we can use the vertex formula.
Step 1: Identify the coefficients of the quadratic equation:
a = 1, b = 8, c = 12
Step 2: Find the x-coordinate of the vertex using the vertex formula:
x = -b / (2a)
x = -8 / (2 * 1) = -4
Step 3: Substitute the value of x into the original function to find the y-coordinate of the vertex:
f(-4) = (-4)^2 + 8(-4) + 12 = 16 - 32 + 12 = -4
Therefore, the vertex of the function is (-4, -4).
Since the coefficient of the x^2 term is positive (a = 1), the graph opens upwards and the vertex (-4, -4) is the minimum point.
So, the value of x at which the minimum occurs is -4 and the point is a minimum.
Therefore, the answer is option (c) -4; minimum.
3. To find the maximum monthly ink revenue for the given revenue function R(x) = -x^2 + 12x + 13, we can use the vertex formula similar to question 2.
Step 1: Identify the coefficients of the quadratic equation:
a = -1, b = 12, c = 13
Step 2: Find the x-coordinate of the vertex using the vertex formula:
x = -b / (2a)
x = -12 / (2 * -1) = 12 / 2 = 6
Step 3: Substitute the value of x into the original function to find the y-coordinate of the vertex:
R(6) = -(6)^2 + 12(6) + 13 = -36 + 72 + 13 = 49
Therefore, the vertex of the function is (6, 49).
Since the coefficient of the x^2 term is negative (a = -1), the graph opens downwards and the vertex (6, 49) is the maximum point.
So, the maximum monthly ink revenue is $49.
Therefore, the answer is option (a) $49.
I hope this helps! Let me know if you have any further questions.
1. To find the maximum and minimum values for Y=28(1.21)^x on the interval 0<x<12, we need to find the critical points of the function. The critical points occur where the derivative of the function is equal to zero or is undefined.
First, let's find the derivative of Y with respect to x:
dY/dx = 28 ln(1.21) (1.21)^x
Setting this derivative equal to zero:
28 ln(1.21) (1.21)^x = 0
Since ln(1.21) is a constant, we can divide both sides by (1.21)^x:
(1.21)^x = 0
This equation has no real solutions, which means there are no critical points within the given interval.
Next, let's check the endpoints of the interval. Plug in x = 0 and x = 12 into the function:
Y(0) = 28(1.21)^0 = 28
Y(12) = 28(1.21)^12 ≈ 275.8
Therefore, the maximum and minimum values for Y on the interval 0<x<12 are 275.8 and 28, respectively.
The correct answer is b) 275.8 and 28.
2. To find the maximum or minimum of f(x) = x^2 + 8x + 12, we can use the vertex formula. The vertex formula for a quadratic function of the form f(x) = ax^2 + bx + c is given by x = -b/ (2a).
In this case, the quadratic function is f(x) = x^2 + 8x + 12. Comparing this to the form ax^2 + bx + c, we see that a = 1, b = 8, and c = 12.
Using the vertex formula:
x = -8 / (2 * 1) = -8 / 2 = -4
Therefore, the maximum or minimum occurs at x = -4.
To determine if it is a maximum or minimum, we can examine the concavity of the parabola. Since the coefficient of the x^2 term is positive (a = 1), the parabola opens upward, indicating that the vertex is a minimum point.
Therefore, the correct answer is c) -4; minimum.
3. To find the maximum monthly ink revenue for R(x) = -x^2 + 12x + 13, we need to find the vertex of the parabola. The vertex formula for a quadratic function is x = -b / (2a).
In this case, the quadratic function is R(x) = -x^2 + 12x + 13. Comparing this to the form ax^2 + bx + c, we see that a = -1, b = 12, and c = 13.
Using the vertex formula:
x = -12 / (2 * -1) = -12 / -2 = 6
Now, substitute x = 6 into the function to find the maximum monthly ink revenue:
R(6) = -(6)^2 + 12(6) + 13
= -36 + 72 + 13
= 49
Therefore, the maximum monthly ink revenue is $49.
The correct answer is a) $49.