Expressing Combined Area of a square and rectangle

sparklemetink

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Here is the question. A piece of wire 3m long is cut into two pieces. Let x denote length of first piece and 3-x the length of the second. First piece is bent into a square; second into a rectangle. Express combined area of the square and the rectangle as a function of x.

Now, my teacher said that the area of the square is x/4. How did she get that? the last time I checked, the area of a square is s^2.

The area of a rectangle is width times length...so this is all I have.

? = x(3-x)
? = 3x-x^2
I just can't figure out the rest.

Help would be appreciated, thanks.

Karen
 
sparklemetink said:
Here is the question. A piece of wire 3m long is cut into two pieces. Let x denote length of first piece and 3-x the length of the second. First piece is bent into a square; second into a rectangle. Express combined area of the square and the rectangle as a function of x.

Now, my teacher said that the area of the square is x/4. How did she get that? the last time I checked, the area of a square is s^2.

The side length of the square is s=x4\displaystyle s=\frac{x}{4} and the area is s2\displaystyle s^{2}.

A square is a rectangle with equal side lengths. So, the side lengths of the rectangle could be 3x4\displaystyle \frac{3-x}{4} and the area

(3x4)2\displaystyle \left(\frac{3-x}{4}\right)^{2}
 
Ok. I think that now makes sense. So, I worked out the problem and the book's answer and my answer is not the same. I get...9-16x+x^2/16
The answer in the book is...17/144x^2-1/3x+1/2 Huh, I don't get it...Sorry. :(
 
Oh, they're using the area of the rectangle as (3x)218\displaystyle \frac{(3-x)^{2}}{18}

By breaking the 3-x length of wire up into unequal size pieces for the rectangle, instead of the same as in the square.

That means the rectangle has side length 3x6\displaystyle \frac{3-x}{6} and width 3x3\displaystyle \frac{3-x}{3}

Therefore, the entire length of the wire for the rectangle portion is the perimeter of the rectangle being

2(3x)6+2(3x)3=3x\displaystyle \frac{2(3-x)}{6}+\frac{2(3-x)}{3}=3-x

Total Area=(3x)218+(x4)2=17x214413x+12\displaystyle \text{Total Area}=\frac{(3-x)^{2}}{18}+(\frac{x}{4})^{2}=\frac{17x^{2}}{144}-\frac{1}{3}x+\frac{1}{2}
 
sparklemetink, in the original problem was the length of the rectangle supposed\displaystyle sparklemetink, \ in \ the \ original \ problem \ was \ the \ length \ of \ the \ rectangle \ supposed

to be twice its width?\displaystyle to \ be \ twice \ its \ width?
 
Karen, you might detest word problems, but if you are unable to describe a\displaystyle Karen, \ you \ might \ detest \ word \ problems, \ but \ if \ you \ are \ unable \ to \ describe \ a

problem you want help on succintly, people might start detesting you.\displaystyle problem \ you \ want \ help \ on \ succintly, \ people \ might \ start \ detesting \ you.
 
Psquare = 4s, s denotes a side, now 4s = x      s = x4.\displaystyle P_{square} \ = \ 4s, \ s \ denotes \ a \ side, \ now \ 4s \ = \ x \ \implies \ s \ = \ \frac{x}{4}.

Hence, Asquare = s2 = (x4)2 = x216.\displaystyle Hence, \ A_{square} \ = \ s^2 \ = \ \bigg(\frac{x}{4}\bigg)^2 \ = \ \frac{x^2}{16}.

Prectangle = 2w + 2l = 3x, l= length and w = width of rectangle.\displaystyle P_{rectangle} \ = \ 2w \ + \ 2l \ = \ 3-x, \ l = \ length \ and \ w \ = \ width \ of \ rectangle.

Now, we were given (somewhat belatedly) that the length is twice the width or l = 2w.\displaystyle Now, \ we \ were \ given \ (somewhat \ belatedly) \ that \ the \ length \ is \ twice \ the \ width \ or \ l \ = \ 2w.

Ergo, Prectangle = 2w+2(2w) =3x, 6w = 3x      w = 3x6 and\displaystyle Ergo, \ P_{rectangle} \ = \ 2w+2(2w) \ =3-x, \ 6w \ = \ 3-x \ \implies \ w \ = \ \frac{3-x}{6} \ and

l = 3x3.\displaystyle l \ = \ \frac{3-x}{3}.

Now Arectangle = lw = (3x3)(3x6) = 96x+x218\displaystyle Now \ A_{rectangle} \ = \ lw \ = \ \bigg(\frac{3-x}{3}\bigg)\bigg(\frac{3-x}{6}\bigg) \ = \ \frac{9-6x+x^2}{18}

Therefore, putting it all together, we have:\displaystyle Therefore, \ putting \ it \ all \ together, \ we \ have:

Total area = x216+96x+x218 = 17x248x+72144 or\displaystyle Total \ area \ = \ \frac{x^2}{16}+\frac{9-6x+x^2}{18} \ = \ \frac{17x^2-48x+72}{144} \ or

A(x) = 17x2144x3+12, QED.\displaystyle A(x) \ = \ \frac{17x^2}{144}-\frac{x}{3}+\frac{1}{2}, \ QED.

Note: Without the "belator", I or anyone else, would be totally at sea.\displaystyle Note: \ Without \ the \ "belator", \ I \ or \ anyone \ else, \ would \ be \ totally \ at \ sea.
 
I just want to appologize to everybody. I was only thinking of myself. I'm very new to these forums. Please forgive me. Next time I will type the problem directly from the textbook.
 
I,myself forgive you Karen.\displaystyle I, myself \ forgive \ you \ Karen.

Post Script: An another thing, I dont detest you.\displaystyle Post \ Script: \ An \ another \ thing, \ I \ don't \ detest \ you.
 
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