Posted by **Jeanna** on Thursday, August 6, 2009 at 11:23pm.

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 -
**Damon**, Friday, August 7, 2009 at 3:11am
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

- Algebra Word Problem Help -
**Jeanna**, Friday, August 7, 2009 at 10:09am
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!

- Algebra Word Problem Help -
**Marth**, Friday, August 7, 2009 at 11:19am
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

## Answer This Question

## Related Questions

- Algebra - A gardener has 60 feet of fencing with which to enclose a garden ...
- Algebra Word Problem Help - A gardener has 40 feet of fencing with which to ...
- Algebra Word Problem Help - I got 2 more intense problems like this. Can someone...
- Algebra Help - I got 2 more intense problems like this. Can someone show me the ...
- Algebra Word Problem Help - A gardener has 40 feet of fencing with which to ...
- Algebra - A builder has 80 feet of fencing to create an enclosure adjacent to a ...
- Math, word problem - A gardener needs to enclose three adjacent rectangular ...
- Algebra - Algebra Word Problems: 1. The length of the floor of a one-storey ...
- college algebra word problem - Here's the problem I was asked: You are visiting ...
- word problem? - A rectangular lot is to be bounded by a fence on three sides and...

More Related Questions