1. find the absolute max and min values of the function, if they exist over the
ID: 2836919 • Letter: 1
Question
1. find the absolute max and min values of the function, if they exist over the indicated interval.
a) f(x)= 12-6x^2 +x^2 defined on [-1,5]
b) f(x)= 4x/ x^2 +4 defined on [-1, 3]
c) f(x)= x^2 +250/x defined on [ 0, infinity]
Q#3 A rancher wants to enclose a rectangular area next to a wall. There are 480 yd of
fencing available. What is the largest total area that can be enclosed?
Q#4 A soup company is constructing an open-top, square-based, rectangular metal tank
that will have a volume of 64 ft^3 . What dimensions yield the minimum surface area? What
is the minimum surface area?
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1. find the absolute max and min values of the function, if they exist over the indicated interval. a) f(x)= 12-6x^2 +x^2 defined on [-1,5] b) f(x)= 4x/ x^2 +4 defined on [-1, 3] c) f(x)= x^2 +250/x defined on [ 0, infinity] Q#2 For a dosage of x cubic centimeters (cc) of a certain drug, the resulting blood pressure B is approximated by B(x) = 0.036x ^2- 0.32x^3 where 0Explanation / Answer
Question 1
a) f'(x) = -10x
Critical points are at f'(x) = 0 --> this happens when x = 0
Find f'(x) values at your endpoints and critical points to find absolute minimum and maximum:
So, from this table we can see that our absolute maximum is 12 at x = 0 because this is the greatest value of f'(x).
Our absolute minimum is -113 at x = 5, since this is the lowest value.
(Do the same for parts b and c--> 1. find the derivative, 2. find your critical point(s), 3. Plug your endpoints and critical points into f'(x) and see which values are greatest and smallest. These are abs. max and min)
For question 2, it is the exact same process. Find B'(x). Find the critical points of B'(x). Plug in endpoints (0 and 0.18) and critical points into B'(x). The greatest value of B'(x) will be the max. blood pressure, and that x will be the dosage at which it occurs. The smallest value of B'(x) will be the min. blood pressure, and that x will be the dosage at which it occurs.
For Question 3, you only need to find the maximum.
You know that the area is W*L.
The 480 yards is equal to 2W + L. SO, this means that L = 480-2W
A(area) = W*L = (480-2W)W = 480W-2(W^2)
Then, dA/dW = 480-4W
If you set dA/dW = 0, you can find W and L.
W = 480/4 = 120
L = 480 -2W = 480-2(120) = 240
A = W*L = (120)(240) = 28800 yd^2
Problem 4 is similar to problem 3, except we are finding the minimum and using the volume.
V = (X^2)*Y = 64 ft^3 X^2 because it is a square base, so the sides are equal
Y = 64/(X^2)
A = X^2 + 4XY Surface area
Substitute Y --> A = X^2 + 4X(64/(X^2)) = X^2 + (256 X/X^2)
A = X^2 + 256/X
Find A': 2X - 256/(X^2)
Find the critical points by setting A' = 0 --> X^3 = 128
X = cube root of 128 = 5.04
Y = 64/(X^2) = 64/(5.04)^2 = 64/25.4 = 2.52
Surface area is minimized when height is 1/2 the base length.
x -1 0 5 f'(x) 7 12 -113Related Questions
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