# Need help with the end of this problem

• January 24th 2013, 09:17 AM
skweres1
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!
• January 24th 2013, 10:14 AM
mathamagician77
Re: Need help with the end of this problem
Everything looks right. 11.51 by 6.51 are your new dimensions.
• January 24th 2013, 10:22 AM
Soroban
Re: Need help with the end of this problem
Hello, skweres1!

Quote:

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

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

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

I assume you made a sketch.

Code:

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