Question 1185499
.
A factory is to be built on a lot measuring 270 ft by 360 ft. A local building code specifies that 
a lawn of uniform width and equal in area to the factory must surround the factory.

What must the width of the lawn be? ?

If the dimensions of the factory are A ft by B ft with A<B, then A= ?
 and B= ?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The answer for the uniform width  &nbsp;180 ft, &nbsp;which @josgarithmetic derived in his post,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;is &nbsp;INCORRECT &nbsp;and &nbsp;TOTALLY &nbsp;ABSURDIST, &nbsp;since it does nor leave a room for the factory inside this lot.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to bring a correct solution and correct answer.



<pre>
The outer dimensions of the lot are 270 ft by 300 ft.


The area of the lot is  270 x 360 = 97200 square feet.


The area of the factory must be half of it, i.e. 48600 square feet.


If u is the uniform width of the lawn, then the equation for the factory area is


    (270-2u)*(360-2u) = 48600.


Simplify and reduce to the standard quadratic equation form


    4u^2 - 2*270u - 2*360u + 97200 = 48600

    4u^2 - 1260u + 48600 = 0

     u^2 -  315u + 12150 = 0


Use the quadratic formula


    {{{u[1,2]}}} = {{{(315 - sqrt(315^2 - 4*12150))/2}}} = {{{(315 +- 225)/2}}}.


Now, even by disarmed eyes, it is clear that  {{{u[1]}}} = {{{(315+225)/2}}} = 270 is inappropriate solution,
since it is tooooo big, not leaving space/room for the factory.


The only remaining solution is {{{u[2]}}} = {{{(315-225)/2}}} = {{{90/2}}} = 45 feet.


Thus the uniform width is 45 feet, and the dimensions of the factory should be

    270 - 2*45 = 180 feet  and  360 - 2*45 = 270 feet.    <U>ANSWER</U>


<U>CHECK</U>.  The factory area  180*270  square feet is EXCATLY HALF  of the lot area  360*270 square feet.  ! Correct !
</pre>

Solved.



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Now @josgarithmetic will be feverishly re-write his post, &nbsp;redoing calculations, &nbsp;to get into my numbers.



I observe his activity at this forum during many years.


Yet &nbsp;3-4 years ago, &nbsp;his scores &nbsp;(the percentage of correctly solved problems) &nbsp;were &nbsp;40 - 50%,

and I wrote about it in the forum, &nbsp;calling him &nbsp;"pseudo"-tutor.


Now his scores are about &nbsp;10%, &nbsp;and it is just another category.



I know it very well, because I fixed wrong solutions after him every day several times per day.



The conclusion is: &nbsp;you, &nbsp;the visitor, &nbsp;should not trust &nbsp;NO &nbsp;ONE  &nbsp;problem, &nbsp;solved, &nbsp;answered and presented by him.



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Everything happened with his post as I predicted . . .