Dimensions word problem

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RHSLilSweetie07

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A homeowner wishes to fence in three adjoining garden plots, one for each of her children. If each plot is to be 80 ft in area, and hwe has 88 ft. of fencing material at hand, what dimensions should each plot have?

How do you set up this problem to work it?
 
Are we assuming the plots to be one big rectangle, with two division lines drawn through it, forming three equal regions?

Eliz.
 
Re: Three equal plots

RHSLilSweetie07 said:
yes, three equal plots
Click to expand...
Three equal rectangular and conjoined plots?

I'll guess so.

Draw the rectangle and the inner boundaries. Label the two lengths as "3L" (because the sides are divided into three pieces, for the three plots, with "L" being the length of each individual plot) and the four widths (that is, the ends and the two the inner boundaries) as "w". For an equation relating L and w to the given length of fencing. Solve this equation for "L=", so you have length in terms of width.

Recalling that "area" is "length times width", for the area formula for one of the sub-plots. Plug in your expression (from above) for L, so you have the area formula for one of the sub-plots only in terms of w.

Set this formula equal to 80, and solve.

Eliz.

Edit: Correcting next-to-last paragraph for consistency.
 
Confused

I'm not sure how to set up the formula. Would it be 3L * W = 240, which equals L = 80/w? Then what do I do?
 
You have another equation based on the 88 feet of fencing.
6L+4W=88
Substitute your L=80/W and solve the quadratic for W
 
Substituting L = 80 / W in 6L + 4W = 88 gives you:
6(80 / W) + 4W = 88
solving gives:
W^2 - 22W + 120 = 0
(W - 10)(W - 12) = 0
W = 10 or W = 12 : so you get 2 answers

Are you able to follow that?
If not, better talk to your teacher...
 
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