Maximum Area

[bkellymusic117]

There are two adjacent rectangular fields, 480 meters of fencing, find the maximum area.

Where am I going wrong?

480=3W+2L
L=240-1.5W
A=LW or A = (240-1.5W)W
-1.5W^2 +240W
h = -240/2(-1.5)= 8
240(8)-1.5(8)2= 1824.
ANSWERS:
9200
9400
9600
9800

Someone help please? thanks.

 

[Deleted member 4993]

There are two adjacent rectangular fields, 480 meters of fencing, find the maximum area.

Where am I going wrong?

480=3W+2L
L=240-1.5W
A=LW or A = (240-1.5W)W
-1.5W^2 +240W
h = -240/2(-1.5)= 8
240(8)-1.5(8)2= 1824.
ANSWERS:
9200
9400
9600
9800

Someone help please? thanks.

Your problem statement is incomplete.

Please post the EXACT problem as it was given to you (if possible, with a sketch).

 

[bkellymusic117]

Your problem statement is incomplete.

Please post the EXACT problem as it was given to you (if possible, with a sketch).

4. A school board is preparing plans for a new middle school. Officials plan to enclose two adjacent rectangular playing fields with fencing that is already in storage. If they have 480 meters of fencing available, what is the maximum area that can be enclosed?

A. 9200 m^2 B. 9400 m^2 C. 9600 m^2 D. 9800 m^2


Huettenmueller, Rhonda (2014-01-01). College Algebra DeMYSTiFieD, 2nd Edition (Kindle Locations 5366-5375). McGraw-Hill Education. Kindle Edition.

 

[Quaid]

Where am I going wrong?

480=3W+2L

L=240-1.5W

A=LW or A = (240-1.5W)W

-1.5W^2 +240W

h = -240/(2(-1.5)) = 8

Hi bkelly:

Your errors are shown in red above.

Without those extra grouping symbols, the Order of Operations is not correct.

Also, check your arithmetic because 8 is not the correct value.

Cheers :cool:

PS: What is h?

 

[JeffM]

There are two adjacent rectangular fields, 480 meters of fencing, find the maximum area.

Where am I going wrong? First, you have not named your variables. That not only makes it difficult to communicate with others, but it also may confuse you by forcing you to keep more things in your memory than you need to.

L = the length of the combined fields.
W = the width of the individual fields.
A = area of combined fields.

480=3W+2L GOOD
L=240-1.5W GOOD
A=LW or A = (240-1.5W)W GOOD
-1.5W^2 +240W Where did the equation go? What you should show is A = -1.5W^2 + 240W.
h = -240/2(-1.5)= 8 It would be nice if you said what h is, but you seem to be going for the formula for determining where the vertex of a parabola is. That's good thinking. However, as quaid said, you failed to use PEMDAS properly and got the arithmetic wrong. One of the NICE things about this kind of multiple choice problem is that (because it does not offer "none of the above" as a choice) you can be sure that you made a mistake somewhere when your answer does not match any of the options. In this case, it was in arithmetic. You should get used to doing a quick check on your arithmetic for reasonableness. 2 times 1.5 is 3. Does it make sense to you that dividing 240 by 3 gets an answer less than 10? This kind of check for reasonableness will prevent a lot of careless errors. This is particularly true when you use a calculator: it is easy to enter the wrong number into a calculator (such as 15 instead of 1.5) and get a silly answer. Calculators may not make mistakes, but the people who use them do.
240(8)-1.5(8)2= 1824.
ANSWERS:
9200
9400
9600
9800

Someone help please? thanks.

I want to clarify something I said above in blue. Back when I was a student, we used slide rules, not calculators. A slide rule did not tell you where the decimal point was in an answer. So you had to to do a very crude estimate in your head to know what the order of magnitude of the answer should be. The technology forced you to make a reasonableness check. The world of calculators does not force you to do that; nevertheless it is a good habit to acquire.