Question

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)

Answer 1

`"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."`

Answer 2

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]}