# Algebra Word Problem Help

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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!

• Algebra Word Problem Help - ,

If each side is x, then the perimeter=2x+L, or L= 40-2x (remember this is only three sides).
Area= x*(40-2x) so as I read B, that is correct.

all else correct.

• Algebra Word Problem Help - ,

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

• Algebra Word Problem Help - ,

(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=40-20
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

• Algebra Word Problem Help - ,

(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=40-20
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?

• Algebra Word Problem Help - ,

JEnna, did I not answer the area? x(40-2x)=-2x^2+40x
What is it you do not understand here?

• Algebra Word Problem Help - ,

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?

• Algebra Word Problem Help - ,

yes. Yes, yes. Have ye no Faith?

• Algebra Word Problem Help - ,

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