Help with Algebra word problem

[jam217]

My son has a word problem for math and we haven't been able to figure it out.

"I have hired a landscape contractor to make a cement border of uniform width around my pool. My pool if 10 feet long and 6 feet wide. The amount I have paid the contractor is enough to purchase 216 square yards of materials. Will this be enough to make a uniform border? If so, how wide will the border be?"

Can someone give us direction on how to solve please?

 

[kaytieeldr]

What's the material?

It does matter - are we talking about a wooden fence, or brick, or stones and cement...? The actual width/depth of the wall itself matters. Also, is that the entire problem or is there any other information available?

You CAN build a wall around your pool; whether the wall will be sensible or not is another story entirely.

ETA: Okay, I see now it's cement. Hold on.

 

[damo]

Let the width of the border be x.
Then the width of the entire area is 10 + 2x.
We will have a top and bottom border with width x and length 10 +2x. The area of these borders will be 2(x)(10+2x)

The two centre borders with also have width x but length will just be the same as the pool which is 6. The area of these parts of the border will be 2(x)(6)

The total area of the border is 2x(10+2x) + 12x must be less than or equal to 216 sq. yards or 1944 square feet

20x + 4x^2 +12x=1944
4x^2 + 32x-1944=0 ......... divide thru by 4
x^2 +8x- 486 =0

Solve by the quadratic formula to get x to be about 18.4 ft. So the width of the border is 18.4 ft.

The question makes more sense with 216 sq. ft instead of yards.

 

[hopemax]

It is a weird question, because I think cement is something that normally comes in cubic feet not square.

In general, I think this is where drawing it out helps. You have a box inside of a larger box. The dimensions of the smaller box are the ones given for the pool: 10 x 6. Since the border needs to be uniform, the dimensions of the larger box are (10 + 2X) and (6 + 2X).

Area of the concrete section = (area of the big box) - (area of the small box)

You also have to make sure the units are the same. We have square feet on the right side of the equation, and square yards on the left side. So you have to convert 216 square yards to square feet.

You should end up with a quadratic equation, and then solve for X.

I see damo posted, and I set up the equation a little different than damo, but you end up with the same quadratic equation.

 

[kaytieeldr]

Uniform border = 54 cubic feet of cement for each side
216 square yards = 24 cubic yards = 216 cubic feet
15 x 15 (uniform) border at one foot deep (wide) would allow a border 3.6 feet high on each side.
Border ten feet from pool on each side would be 20 x 12; at the same one foot deep/wide, it would be 3.375 feet high.
Sensible border - one designed to keep people out for safety - should be at least four feet high. Using the same one foot deep (wide) and uniformity requirement, allowing room for a border around the pool, a wall four feet from the pool's edge would be 14 x 10 and would be 4.5 feet high.

Of course, none of these numbers allows for ACCESS to the pool, e.g. a gate!

 

[Micca]

Let the width of the border be x.
Then the width of the entire area is 10 + 2x.
We will have a top and bottom border with width x and length 10 +2x. The area of these borders will be 2(x)(10+2x)

The two centre borders with also have width x but length will just be the same as the pool which is 6. The area of these parts of the border will be 2(x)(6)

The total area of the border is 2x(10+2x) + 12x must be less than or equal to 216 sq. yards or 1944 square feet

20x + 4x^2 +12x=1944
4x^2 + 32x-1944=0 ......... divide thru by 4
x^2 +8x- 486 =0

Solve by the quadratic formula to get x to be about 18.4 ft. So the width of the border is 18.4 ft.

The question makes more sense with 216 sq. ft instead of yards.

Any way I can put $100 on you curing cancer?

 

[damo]

Any way I can put $100 on you curing cancer?

Will paypal work for you?

 

[damo]

Uniform border = 54 cubic feet of cement for each side
216 square yards = 24 cubic yards = 216 cubic feet
15 x 15 (uniform) border at one foot deep (wide) would allow a border 3.6 feet high on each side.
Border ten feet from pool on each side would be 20 x 12; at the same one foot deep/wide, it would be 3.375 feet high.
Sensible border - one designed to keep people out for safety - should be at least four feet high. Using the same one foot deep (wide) and uniformity requirement, allowing room for a border around the pool, a wall four feet from the pool's edge would be 14 x 10 and would be 4.5 feet high.

Of course, none of these numbers allows for ACCESS to the pool, e.g. a gate!

Usually when we do these types questions in algebra, they just assume we are talking a flat surface --- a patio or sidewalk --- no height involved.

 

[Micca]

Will paypal work for you?

Maximum respect for math skills.

 

[kaytieeldr]

Oh, I see! Like a deck sort of thing! Now I get it - thanks, Micca! And here I am, trying to figure out if it's cement, if there's enough to make a structure thick enough to be a wall... and then how to get TO the pool!

Good thing I can deal with being laughed at!

 

[HappyGilmore]

Oh good gracious send me back to 5th grade please. I had trouble with just reading the question... by the time I got to the end I forgot the beginning. My poor kids!! I will be of no help once they get into this level. Humor me and tell me what grade math this is? I pray it is high school but I am thinking it may be more like middle school.

 

[laurie31]

It's either 8th or 9th grade, depending on your school system.

 

[jam217]

Thank you for all your replies and help! It is very much appreciated. This is an algebra II math question for those who were wondering what level.

I am clueless with this stuff so that's why we asked for help. My son wouldn't ask any of his friends for assistance so I suggested the disboards and as usual, you came through! You guys are great!

 

[themilesfamily]

I'm confused.

I'm with you up to:

x2+8x-486<0 (a quadratic inequality -- because the size of the border should be less than or equal to 216 sq. yards/1944 sq. ft -- you can't go over the amount of materials you've been given.)

but how does that factor down into 18.4 feet?

If the OP typed it wrong and it was 216 sq. feet, I can follow you to:
x2+8x-54<0

But again, I can't figure out how to factor that out and solve for X.

Can someone come back and tell me how to figure out a quadratic equation or quadratic inequality when the numbers don't factor out evenly?

 

[damo]

I'm confused.

I'm with you up to:

x2+8x-486<0 (a quadratic inequality -- because the size of the border should be less than or equal to 216 sq. yards/1944 sq. ft -- you can't go over the amount of materials you've been given.)

but how does that factor down into 18.4 feet?

If the OP typed it wrong and it was 216 sq. feet, I can follow you to:
x2+8x-54<0

But again, I can't figure out how to factor that out and solve for X.

Can someone come back and tell me how to figure out a quadratic equation or quadratic inequality when the numbers don't factor out evenly?

You have to use the quadratic formula

x=[-b+-SQRT(b^2-4ac)]/2a where the quadratic equation is ax^2 +bx +c = 0