Question 390023
A rectangle area:

{{{A = ab}}}....where {{{a}}} and {{{b}}} are length and width

given:

{{{A =575m^2}}} and {{{100m}}} for fencing which is equal to perimeter {{{P = 2(a+b)}}}

to find: {{{a}}} and {{{b}}}

use

{{{ab =575m^2}}} and {{{100m = 2(a+b)}}} and solve the system


{{{ab =575m^2}}}...solve for {{{a}}}

{{{a =575m^2/b}}}........substitute in {{{100m = 2(a+b)}}} and find {{{b}}}


{{{cross(100)50m = cross(2)((575m^2/b)+b)}}} 


{{{50m = (575m^2/b)+b}}}...both sides multiply by {{{b}}}


{{{50m *b= b(575m^2/b)+b*b}}}


{{{50m *b= cross(b)(575m^2/cross(b)) +b^2}}}


{{{50m *b= 575m^2 +b^2}}}


{{{50m *b- b^2= 575m^2}}}



{{{- b^2 + 50m *b -575m^2 =0}}}...solve for {{{b}}}


*[invoke quadratic_formula -1, 50, -575, "x"]

one solution

{{{b=25-5sqrt(2)=25-5*1.41= 25-7.05= 17.95m}}}


{{{b= 17.95m}}}

other solution

{{{b=25+5sqrt(2)=25+5*1.41= 25+7.05= 32.05m}}}

{{{b= 32.05m}}}

now find {{{a}}}

one solution

{{{a =575m^2/17.95m}}}


{{{a =32.03m}}}

other solution

{{{a =575m^2/32.05m}}}

{{{a =17.94m}}}


so, dimensions are:


{{{a =32.03m}}}


{{{b= 17.95m}}}


or


{{{a =17.94m}}}

{{{b= 32.05m}}}


check:

{{{100m = 2(a+b)}}}

{{{100m = 2(32.03m+17.94m)}}}

{{{100m = 2(49.97m)}}}..round decimal number to nearest whole number


{{{100m = 2(50m)}}}


{{{100m = 100m}}}