Algebra Word Problem Help
posted by Jeanna on .
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=4020
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!

If each side is x, then the perimeter=2x+L, or L= 402x (remember this is only three sides).
Area= x*(402x) so as I read B, that is correct.
all else correct. 
oops. You erred. the max area occurs when x is 10, and W is 20. You divided wrong (Perimeter is 40, so W cannot be 200)
Max area 200 
(a) 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? (No explanation required) Choice: B?
A. A(x) = –2x2 + 20x
B. A(x) = –2x2 + 40x
C. A(x) = 2x2 – 40x
D. A(x) = x2 – 40x
The area function is a quadratic function and so its graph is a parabola.
(b) Does the parabola open up or down? down?
(c) Find the vertex of the quadratic function A(x). Show work.
(b/2a, f (b/2a))
(0+20)/2 =10
2x^2+40x = 200
Vertex = (10,200)
(d) 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 200 sqft ?
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=4020
y=20
x=10
20*10 = 200 sqft
When the sides perpendicular to the wall have length x = 10ft?
and the side parallel to the wall has length 20ft?.
A = 2x^2+40 x
A = 2(100) + 40(10)
A = 200 
(a) 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? (No explanation required) Choice: B?
A. A(x) = –2x2 + 20x
B. A(x) = –2x2 + 40x
C. A(x) = 2x2 – 40x
D. A(x) = x2 – 40x
The area function is a quadratic function and so its graph is a parabola.
(b) Does the parabola open up or down? down?
(c) Find the vertex of the quadratic function A(x). Show work.
(b/2a, f (b/2a))
(0+20)/2 =10
2x^2+40x = 200
Vertex = (10,200)
(d) 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 200 sqft ?
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=4020
y=20
x=10
20*10 = 200 sqft
When the sides perpendicular to the wall have length x = 10ft?
and the side parallel to the wall has length 20ft?
A = 2x^2+40 x
A = 2(100) + 40(10)
A = 200
Is this all correct now? 
JEnna, did I not answer the area? x(402x)=2x^2+40x
What is it you do not understand here? 
B.A(x) = –2x2 + 40x
(b) Does the parabola open up or down? down?
VERTEX
(b/2a, f (b/2a))
(0+20)/2 =10
2x^2+40x = 200
Vertex = (10,200)
The maximum area is 200 sqft ?
When the sides perpendicular to the wall have length x = 10ft?
and the side parallel to the wall has length 20ft? .
A = 2x^2+40 x
A = 2(100) + 40(10)
A = 200
Is this all correct now? 
yes. Yes, yes. Have ye no Faith?

I do, but I wanted to make sure my calcualtions were accurate and correct. Thanks!