The department of transportation has proposed a new 7-mile stretch of highway. The elevation of the highway x miles along the path is given by E(x)=.004 (x^4-8x^3)+2. Where is the steepest point
that's where E' is a max
E' = .004(4x^3-24x^2) = .008(x^3-6x^2)
E' is max where E" = 0
E" = .008(3x^2-12x) = .024x(x-4)
so, |E'| is max when x=4, as you can see at
http://www.wolframalpha.com/input/?i=.004+%28x%5E4-8x%5E3%29%2B2
To find the steepest point on the highway, we need to determine where the highest elevation occurs. This can be done by finding the maximum point of the elevation function E(x).
To find the maximum point of a function, we can use calculus by finding the derivative of the function and setting it equal to zero. So, let's start by finding the derivative of E(x):
E'(x) = d/dx (.004 (x^4-8x^3)+2)
To calculate the derivative, we need to apply the power rule and constant rule. The derivative of x^4 is 4x^3, and the derivative of -8x^3 is -24x^2. Since the constant 2 doesn't change when taking the derivative, it becomes 0. The derivative simplifies to:
E'(x) = .004 (4x^3 - 24x^2)
Next, we set the derivative equal to zero and solve for x:
0 = .004 (4x^3 - 24x^2)
Simplifying:
0 = 4x^3 - 24x^2
Now, we can factor out an x^2 from both terms:
0 = x^2(4x - 24)
Since a product is equal to zero if and only if at least one of the factors is zero, we set each factor equal to zero:
x^2 = 0 or 4x - 24 = 0
For x^2 = 0, the only solution is x = 0.
For 4x - 24 = 0, we solve for x:
4x = 24
x = 24/4
x = 6
Now we have two possible points to consider: x = 0 and x = 6.
To determine which of these points is the maximum, we can use the second derivative test. This test involves taking the second derivative of the function and evaluating it at each critical point to determine concavity.
The second derivative can be found by taking the derivative of the first derivative:
E''(x) = .004 (12x^2 - 48x)
Evaluating E''(x) at x = 0:
E''(0) = .004 (12(0)^2 - 48(0))
E''(0) = .004 (0 - 0)
E''(0) = 0
Evaluating E''(x) at x = 6:
E''(6) = .004 (12(6)^2 - 48(6))
E''(6) = .004 (432 - 288)
E''(6) = .004 (144)
E''(6) = 0.576
Since E''(0) = 0 and E''(6) = 0.576 > 0, we can conclude that x = 6 corresponds to the steepest point on the highway, as it is where the elevation function changes from concave down to concave up.
Therefore, the steepest point on the proposed 7-mile stretch of highway is located 6 miles along the path.