Expressing Combined Area of a square and rectangle

[sparklemetink]

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

 

[galactus]

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 \(\displaystyle s=\frac{x}{4}\) and the area is \(\displaystyle s^{2}\).

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

\(\displaystyle \left(\frac{3-x}{4}\right)^{2}\)

 

[sparklemetink]

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.

 

[galactus]

Oh, they're using the area of the rectangle as \(\displaystyle \frac{(3-x)^{2}}{18}\)

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 \(\displaystyle \frac{3-x}{6}\) and width \(\displaystyle \frac{3-x}{3}\)

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

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

\(\displaystyle \text{Total Area}=\frac{(3-x)^{2}}{18}+(\frac{x}{4})^{2}=\frac{17x^{2}}{144}-\frac{1}{3}x+\frac{1}{2}\)

 

[sparklemetink]

Ok...I think I finally get it. Thank you. I so detest word problems.

 

[BigGlenntheHeavy]

\(\displaystyle sparklemetink, \ in \ the \ original \ problem \ was \ the \ length \ of \ the \ rectangle \ supposed\)

\(\displaystyle to \ be \ twice \ its \ width?\)

 

[sparklemetink]

Oh..look at that...yes! It says the width is half the length.

 

[BigGlenntheHeavy]

\(\displaystyle Karen, \ you \ might \ detest \ word \ problems, \ but \ if \ you \ are \ unable \ to \ describe \ a\)

\(\displaystyle problem \ you \ want \ help \ on \ succintly, \ people \ might \ start \ detesting \ you.\)

 

[BigGlenntheHeavy]

\(\displaystyle P_{square} \ = \ 4s, \ s \ denotes \ a \ side, \ now \ 4s \ = \ x \ \implies \ s \ = \ \frac{x}{4}.\)

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

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

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

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

\(\displaystyle l \ = \ \frac{3-x}{3}.\)

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

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

\(\displaystyle Total \ area \ = \ \frac{x^2}{16}+\frac{9-6x+x^2}{18} \ = \ \frac{17x^2-48x+72}{144} \ or\)

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

\(\displaystyle Note: \ Without \ the \ "belator", \ I \ or \ anyone \ else, \ would \ be \ totally \ at \ sea.\)

 

[sparklemetink]

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.

 

[BigGlenntheHeavy]

\(\displaystyle I, myself \ forgive \ you \ Karen.\)

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