Thread: Need help with the end of this problem

1. Need help with the end of this problem

In order to save space, both the length and width of an advertisement in a newspaper had ot be decreased by the same amount in order to cut its area in half. The original advertisement had a length of 10 centimeters and a width of 15 centimeters. What are the new dimensions of the advertisement?

My work:
(15-x)910-x)=75
150-15x-10x+x^2=75
x^2-25x+75=0
Then set it up in a quadratic and got 21.51 and 3.49. Would that be my answers or would I need to subtract those by 15-x and 10-x to find my new dimensions? If so would it be 15-3.49 and 10-3.49?

Thanks!

2. Re: Need help with the end of this problem

Everything looks right. 11.51 by 6.51 are your new dimensions.

3. Re: Need help with the end of this problem

Hello, skweres1!

In order to save space, both the length and width of an advertisement in a newspaper
had to be decreased by the same amount in order to cut its area in half.

My work:
. . $\displaystyle (15-x)(10-x)\:=\: 75 \quad\Rightarrow\quad x^2-25x+75\:=\:0$ . Good!

Then set it up in a quadratic and got $\displaystyle 21.51$ and $\displaystyle 3.49$. . Right!
Would that be my answers? . No, x is the width of the strips trimmed off.
. . or would I need to subtract those by $\displaystyle 15-x$ and $\displaystyle 10-x$ to find my new dimensions? . Yes!

If so, would they be $\displaystyle 15-3.49$ and $\displaystyle 10-3.49$? . Correct!

I assume you made a sketch.

Code:
      : - - 10  - - :
- *---*-------- * -
: |///://///////| x
: * - + - - - - * -
: |///:         | :
: |///:         | :
15 |///:         | :
: |///:         |15-x
: |///:         | :
: |///:         | :
: |///:         | :
- *---*---------* -
: x :   10-x  :