Inequality equations

by Erin
(Lakeland, Florida)

The question asks to write and solve an inequality to find the length of a rectangle and to write an inequality to find the area of the rectangle.

It tells that the width is 33 cm and the perimeter is at least 776 cm. How can I do this problem?

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Jan 26, 2015
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starstarstarstarstar 		Here's How You would do it NEW
by: Anonymous

Hi Erin,
What I would do is double the width and subtract it from the total perimeter, then i would take the sum of the last step and divide it by two,this will give you the length. Once you have the side length you can pretty much do anything.

Expression:

1/2 (p-2w)

Where:
p = the perimeter
w = width of rectangle
Jan 20, 2011
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starstarstarstarstar 		thanks
by: Erin

hi thank you so much this helped alot!
Jan 20, 2011
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starstarstarstarstar 		Inequality Equations
by: Karin

Hi Erin,

I know you are asked to do a lot for this problem.

Let's start with the perimeter. We know that the perimeter formula is:

2w +2l = p

The width is 33 cm and we don't know the length. We know the perimeter is at least (means greater than or equal to) 776 cm.

So, we know:
2(33) + 2L > (or equal to) 776

66 + 2L > (or equal to ) 776

Now we want to solve for L.

Subtract 66 from both sides.

66-66 + 2L > (or equal to ) 776-66

2L > (or equal to) 710

Now divide both sides by 2.

2L/2 >(or equal to) 710/2

L >(or equal to) 355.

This means that the length is 355 or larger.

So, now we need to write an inequality to solve for the area.

Since we know the width is 33cm and the length is atleast 355, we can multiply 33 x 355 to find the minimum area.

33 x 355 = 11715. This means that the area is at least 11715 cm (squared).

So, the formula for Area is L x W

So, L x 33 > (or equal to) 11715

Or this could be written as:

33L > (or equal to) 11715.

I hope this helps. I can't write the greater than or equal to symbol in this application. That is why I put that phrase in parenthesis.

Karin

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