Question

How do I solve this?

Answer 1

`\sf\\P=2L+2W=25 \sf\\A=LW=36\\\\Find\ the\ value\ of\ one\ of\ the\ variables\ in\ terms\ of\ the\ other. \sf\\36=LW\\W= (36)/(L)\\\\Substitute.\\P=25\\=2L+2W\\=2L+2( (36)/(L) )\\=2L+ (72)/(L) \\\\Make\ them\ have\ a\ common\ denominator.\\2L+(72)/(L)\\= (2L)/(1) +(72)/(L)\\=(2L^2)/(L)+ (72)/(L) \\= (2L^2+72)/(L) \\\\Multiply\ L\ on\ the\ other\ side.\\25L=2L^2+72\\0=2L^2-25L+72\\\\Use\ the\ quadratic\ formula.\\ L=\frac{-b+/- \sqrt{b^(2)-4ac} }{2a}\\ax^2+bx+c`

`a=2\\b=-25\\c=72\\\\ L=\frac{-(-25) +/- \sqrt{(-25)^(2)-4(2)(72)} }{2(2)} \\ =(25+/- √(625-576) )/(4) \\ =(25+/- √(49) )/(4)\\ =(25(+/-)7)/(4) \\\\\sf\ We\ now\ have\ two\ options.\\(1) L=(25+7)/(4)= (32)/(4) =8\\(2)L=(25-7)/(4)= (18)/(4) =4.5\\\\\sf\ Either\ W\ is\ 8\ and\ L\ is\ 4.5\ or\ W\ is\ 4.5\ and\ L\ is\ 8.\ It\ doesn't\ matter.\\\\{\boxed{The\ dimensions\ are\ 8\ cm\ by\ 4.5\ cm.}`

Answer 2

area=LW
perimiter=2(L+W)

aera=36
P=25

36=LW
25=2(L+W)


25=2(L+W)
divide both sides by 2
12.5=L+W
minus W
12.5-W=L

sub for L
36=W(12.5-W)
36=12.5W-W^2
minus (12.5W-W^2) both sides
0=W^2-12.5W+36
use quadratic formula

if you have
ax^2+bx+c=0
x=
`\frac{-b+/- \sqrt{b^(2)-4ac} }{2a}`

a=1
b=-12.5
c=36

W=
`\frac{-(-12.5)+/- \sqrt{(-12.5)^(2)-4(1)(36)} }{2(1)}`
W=
`(12.5+/- √(156.25-144) )/(2)`
W=
`(12.5+/- √(12.25) )/(2)`
aprox
W=8 or 4.5

sub
12.5-W=L


12.5-8=L=4.5
12.5-4.5=L=8
either way

the dimentions are 4.5cm by 8 cm