A gardener has 40 feet of fencing with which to enclose a garden adjacent to a long existing wall. The gardener will use the wall for one side and the available fencing for the remaining three sides.
If the sides perpendicular to the wall have length x feet, which of the following (A, B, C, or D) represents the area A of the garden?
A. A(x) = –2x^2 + 20x
B. A(x) = –2x^2 + 40x
C. A(x) = 2x^2 – 40x
D. A(x) = x^2 – 40x
I say it is B. A(x) = –2x^2 + 40x, any thoughts?
The area function is a quadratic function and so its graph is a parabola.
Does the parabola open up or down? I say down, am I correct?
Find the vertex of the quadratic function:
(-b/2a, f (-b/2a))
(0+20)/2 =10
-2x^2+40x = 200
Vertex = (10,200)
Is my vertex correct?
Use the work in the previous parts to help determine the dimensions of the garden which yield the maximum area, and state the maximum area. (Fill in the blanks below. Include the units of measurement.)
The maximum area is 2000 sqft. Is this accurate?
-2x^+40x
After differentiating and equating to 0
= -4x + 40=0
Putting this value in area function..
Or.
in the perimeter function.
4x=40
2x+y=40
Put x=10
y=40-20
y=20
x=10
200*10 = 2000 sqft
When the sides perpendicular to the wall have length x = 10ft
and the side parallel to the wall has length 200ft
Are these calculations correct based on my original answer of
B. A(x) = –2x^2 + 40x
Any feedback is very much appreciated! Thanks!
Your vertex of the A(x) area function is (10, 200), so the maximum area is 200 square feet.
x is the length of the sides perpendicular to the wall, and 200/x is the length of the side parallel to the wall.
x=10, so the length of the sides perpendicular to the wall is 10ft, and the length of the side parallel to the wall is 20ft
all correct until you maximize
vertex at (10,200) is right.
That is the top of the parabola, maximum
A = -2x^2+40 x
A = -2(100) + 40(10)
A = 200
or using calculus
dA/dx = 0 = -4x + 40
x = 10
Then the same again
Clearly the side opposite the wall can not be 200 feet if you only have 40 feet of fence
So then everything is correct? What do I need to change?
When the sides perpendicular to the wall have length x = 10ft?
and the side parallel to the wall has length 200ft Or should this be 20ft?
The maximum area is 2000 sqft?
Thanks!
Your calculations are mostly correct. Let's go through them step by step to confirm:
1. You correctly identified B. A(x) = –2x^2 + 40x as the area function. This represents the area of the garden based on the length x of the sides perpendicular to the wall.
2. You are correct that the parabola opens downward because the coefficient of the x^2 term is negative (-2x^2).
3. The vertex formula you used is correct. By plugging in the values of a and b into the formula (-b/2a), you correctly found the x-coordinate of the vertex to be 10. However, the y-coordinate of the vertex should be found by substituting this value (x = 10) into the area function (A(x)).
-2(10)^2 + 40(10) = -200 + 400 = 200
So the correct vertex coordinates are (10, 200), as you mentioned.
4. To find the dimensions that yield the maximum area, you can consider the value of x at the vertex (x = 10) along with the perimeter equation you provided.
The perimeter equation, 2x + y = 40, represents the fact that the sum of all three sides (2 sides perpendicular to the wall and the remaining side) should equal the total fencing length of 40 feet.
Plugging in x = 10:
2(10) + y = 40
20 + y = 40
y = 40 - 20
y = 20
So the dimensions of the garden that yield the maximum area are x = 10ft and y = 20ft. The side parallel to the wall is 20ft, not 200ft.
The maximum area is obtained by substituting these dimensions into the area function:
A(10) = -2(10)^2 + 40(10) = -200 + 400 = 200 sqft
So the correct maximum area is 200 sqft, not 2000 sqft as you mentioned.
Overall, your calculations and reasoning are accurate. Just make sure to double-check the final dimensions and maximum area.