A 20 cm length of string is cut into two pieces. One of the pieces is used to form a perimeter of a square?

A 20 cm length of string is cut into two pieces. One of the pieces is used to form a perimeter of a square. The other piece is used to form the perimeter of a right angled isosceles triangle. Find the maximum and minimum total area that can be formed. (THIS QUESTION CAN'T USE CALCULUS)

2 Answers
Feb 25, 2018

#"Minimum total area = 10.175 cm²."#
#"Maximum total area = 25 cm²."#

Explanation:

#"Name x the length of the piece to form a square."#

#"Then the area of the square is "(x/4)^2"."#
#"The perimeter of the triangle is "20-x"."#
#"If y is one of the equal sides of the triangle, then we have"#
#2*y+sqrt(y^2+y^2) = 20-x#
#=> y*(2+sqrt(2)) = 20-x#
#=> y = (20-x)/(2+sqrt(2))#
#=> area = y^2/2 = (20-x)^2/((4+2+4 sqrt(2))*2)#
#= (20-x)^2 / (12+8 sqrt(2))#

#"Total area = "(x/4)^2 + (20-x)^2 / (12+8 sqrt(2))#
# = x^2/16 + x^2/(12+8 sqrt(2)) - 40 x / (12+8 sqrt(2)) + 400/(12+8sqrt(2))#
#= x^2 (1/16 + 1/(12+8sqrt(2))) - (40/(12+8sqrt(2))) x + 400/(12+8sqrt(2))#

#"This is a parabole and the minimum for a parabole"#
#a x^2 + b x + c = 0 " is "x = -b/(2*a)", if a > 0."#
#"The maximum is "x->oo", if a > 0."#

#"So the minimum is"#
#x = 40/(12+8sqrt(2))/(1/8+1/(6+4sqrt(2)))#
#= 40/(12+8sqrt(2))/((6+4sqrt(2)+8)/(8(6+4sqrt(2))))#
#= 160/(14 + 4 sqrt(2))#
#= 160*(14-4 sqrt(2))/(196-32)#
#= (160/164)*(14-4*sqrt(2))#
#= (80/41)*(7-sqrt(8))#

#= 8.13965 " cm"#
#=> " Total area = "10.175" cm²."#

#"The maximum is either x=0 or x=20."#
#"We check the area :"#
#"When "x=0 => " area = "400/(12+8sqrt(2)) = 17.157" cm²"#
#"When "x=20 => " area = "5^2 = 25 " cm²"#
#"So the maximum total area is 25 cm²."#

Feb 25, 2018

The minimum area is #10.1756# and maximum is #25#

Explanation:

The perimeter of a right angled isosceles triangle of side #a# is #a+a+sqrt2a=a(2+sqrt2)# and its area is #a^2/2#,

Let one piece be #x# cm. from which we form a right angled isosceles triangle. It is apparent that side of the right angled isosceles triangle would be #x/(2+sqrt2)# and its area would be

#x^2/(2(2+sqrt2)^2)=x^2/(2(6+4sqrt2))#

= #(x^2(6-4sqrt2))/(2(36-32))=(x^2(3-2sqrt2))/4#

The perimeter of other portion of string which forms a square is #(20-x)# and as side of square is #(20-x)/4# its area is #(20-x)^2/16# and total area #T# of the two is

#T=(20-x)^2/16+(x^2(3-2sqrt2))/4#

= #(400-40x+x^2)/16+(x^2(3-2sqrt2))/4#

= #25-(5x)/2+x^2(1/16+(3-2sqrt2)/4)#

Observe that #3-2sqrt2>0#, hence coefficient of #x^2# is positive and hence we will have a minima and we can write #T# as

#T=0.1054x^2-2.5x+25#

= #0.1054(x^2-23.7192x+(11.8596)^2)+25-0.1054xx(11.8596)^2#

= #0.1054(x-11.8596)^2+10.1756#

As #0.1054(x-11.8596)^2# is always positive, we have minimum value of #T# when #x=11.8596#.

Observe that theoritically there is no maxima for the function, but as value of #x# lies between #[0,20]#, and when #x=0#, we have #T=0.1054(0-11.8596)^2+10.1756#

= #0.1054xx11.8596^2+10.1756=25#

and when #x=20# when #T=0.1054(20-11.8596)^2+10.1756#

= #0.1054xx8.1404^2+10.1756=17.16#

and hence maxima is #25#

graph{25-(5x)/2+x^2(1/16+(3-2sqrt2)/4) [-11.92, 28.08, -0.96, 19.04]}