Need Help with Quadratics question

[hamid14]

I can't figure out 2 of these quadratics word problems on a review sheet which I am studying from. Does anyone know how to solve this?

The first problem:

A framed picture has a border the same area as the picture. The picture has dimensions of 17 by 12 cm. Find the width of the border.

Answer is supposed to be 2.9cm.

The second problem:

A rectangular plot of land is to be enclosed by a fence and divided into three identical rectangular fields. To do this, Diane puts two parallel fences down the field. If fencing down the middle costs $8/m and fencing around the area costs $12/m

a) What is the max area that can be enclosed for $1920?
b) What are the dimensions of each of the 3 fields?

I will be trying to figure these out, so if you can ge me some tips on how to solve these I'd appreciate it. It's just review, it isnt homework.

[DtY]

For the first one:

To calculate the area of the border, subtract the area of the frame:

a(w) = (17+w)*(12+w) - 12*17

Is the function that will give you the area for a given border width. The area of the picture is 12*17, you need to find what w will give a(w) equal to this.

[hamid14]

i had the same equation written and i used quadratic formula but the answers were wrong.

its this right?

204 = (17+w) * (12+w) - 204

i get 5.8 and the other number is negative. am i doing something wrong?

[andrew.]

The "w" in DtY's post is double the actual width because it accounts for both the top and bottom of the frame. Either divide the answer in half at the end, or do it like this:

204 = (17 + 2w)(12 + 2w) - 204

Now "w" will be the width of each side of the border. I believe this is the answer you're looking for.

[hamid14]

ok i had that formula too, but i must have multiplied out the terms incorrectly. any tips for the second question?

[DtY]

andrew. @ Thu Oct 07, 2010 6:25 pm wrote:

The "w" in DtY's post is double the actual width because it accounts for both the top and bottom of the frame. Either divide the answer in half at the end, or do it like this:

204 = (17 + 2w)(12 + 2w) - 204

Now "w" will be the width of each side of the border. I believe this is the answer you're looking for.

Thanks for pointing that out, I hadn't noticed that!

As for the second question, I'll try to do my best to not give away too much. If you have a rectangle with the dimensions w*h where 2w*2h=n, where n is constant (the perimeter is always the same), what ratio of w:h will maximize x*y? If you're not sure, pick an arbitrary n, and plot the ratio vs. area graph (it should be a parabola).

I am pretty sure that if you were asked this on a test, you should be expected to know what ratio will maximize area (if you don't, I'm sure it will be obvious once you figure it out), and then you just need to figure how big you can make that rectangle to fit within the price restriction.