Marquise has 200200200 meters of fencing to build a rectangular garden.
1 answer:
Answer:
2500 Square meters
Step-by-step explanation:
Given the garden area (as a function of its width) as:
The maximum possible area occurs when we maximize the area. To do this, we take the derivative, set it equal to zero and solve for w.
A'(w)=-2w+100
-2w+100=0
-2w=-100
w=50 meters
Since Marquise has 200 meters of fencing to build a rectangular garden,
Perimeter of the proposed garden=200 meters
Perimeter=2(l+w)
2(l+50)=200
2l+100=200
2l=200-100=100
l=50 meters
The dimensions that will yield the maximum area are therefore:
Length =50 meters
Width=50 meters
Maximum Area Possible =50 X 50 =<u>2500 square meters.</u>
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Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
