Respuesta :

apologiabiology
perimiter=2L+2W

w=x
P=80

80=2L+2x
diviide by 2
40=L+x

40-L=x
40-x=L


the area is found by legnth times width
LW or Lx
for a maximum area, try to get the legnths of the sides the same
therefor
L=W=x for max area
40=x+x
40=2x
divide 2
20=x
max area =legnth times width=20^2=400

note: all squares are rectangles, but not all rectangles are squares


L=40-x
A=(40-x)(x)

max area=400 cm^2

Cricetus

The maximum area will be "400 cm²"

Given:

Perimeter = 80 cm

then,

  • [tex]2(l+x) = 80 \ cm[/tex]

Length,

  • [tex]l = (40-x)cm[/tex]

Area,

  • = [tex]lx[/tex]

        = [tex](40-x)x \ cm^2[/tex]

For, Area to be maximum

→                  [tex]\frac{dA}{dx} =0[/tex]

→ [tex]\frac{d}{dx} (40-x)(x)=0[/tex]

→           [tex]40-2x =0[/tex]

→                   [tex]2x =40[/tex]

→                     [tex]x = \frac{40}{2}[/tex]

→                        [tex]= 20[/tex]

hence,

The maximum area,

= [tex](40-20)20[/tex]

= [tex]20\times 20[/tex]

= [tex]400 \ cm^2[/tex]

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