The area of the garden enclosed by the fencing is
A(x, y) = xy
and is constrained by its perimeter,
P = x + 2y = 200
Solve for x in the constraint equation:
x = 200 - 2y
Substitute this into the area function to get a function of one variable:
A(200 - 2y, y) = A(y) = 200y - 2y²
Differentiate A with respect to y :
dA/dy = 200 - 4y
Find the critical points of A :
200 - 4y = 0 ⇒ 4y = 200 ⇒ y = 50
Compute the second derivative of A:
d²A/dy² = -4 < 0
Since the second derivative is always negative, the critical point is a local maximum.
If y = 50, then x = 200 - 2•50 = 100. So the farmer can maximize the garden area by building a (100 ft) × (50 ft) fence.
Answer:
First we need to calculate the distance between Clifton and Burlington by using Pythagorean theorem:
x² + 65² = 97²
=> x² = 97² - 65²
=> x² = 5184
=> x = √5184 = 72 (m)
The total distance from Aurora to Clifton through Burlington is: 65 + 72 = 137 (m)
We have: 137 - 97 = 40 (m)
So it is 40 m closer to travel from Aurora to Clifton directly than from Aurora to Clifton through Burlington
Answer:
Ryan practiced 33.66 minutes on Wednesday
Step-by-step explanation:
You do 43.2+37.48 and it equals 80.68
Then you do 114.34-80.68 which equals 33.66
Answer:
156.25
Step-by-step explanation:
So, we know that this is a square so we know that each side will equal the same.
So, all we need to do is divide 625 by 4. We will get 156.25.
Hope this helps!
Answer:
x = 3
Step-by-step explanation:
Using the rules of logarithms
ln x = ln y ⇒ x = y
ln e = 1
Given
2 ln e ln 5x = 2 ln 15 ( divide both sides by 2 )
ln e ln 5x = ln 15
ln 5x = ln 15, hence
5x = 15 ( divide both sides by 5 )
x = 3
