Question #40c26
1 Answer
Minimum area of
Explanation:
Let us set up the following variables:
{(t, "Length of a side of the triangle"), (s, "Length of a side of the square"), (A, "Total Area") :}
Our aim is to find
The total perimeter is that of
\ \ \ 10 = 3t+4s
:. s = 1/4(10-3t)
And the total Area is that of an equilibrium triangle and a square:
A = (1/2)(t)(t)sin60^o + (s)(s)
\ \ \ = (1/2)t^2(1/2sqrt(3)) + s^2
\ \ \ = 1/4sqrt(3) \ t^2 + 1/16(10-3t) ^2
We now have the Area,
(dA)/(dt) = (2)(1/4sqrt(3))t+1/16(2)(10-3t)(-3)
\ \ \ \ \ \ = 1/2sqrt(3) \ t-3/8(10-3t)
At a critical point we have
1/2sqrt(3) \ t - 3/8(10-3t) = 0
:. 1/2sqrt(3) \ t -15/4 + 3/4t = 0
:. (1/2sqrt(3) + 3/4)t = 15/4
:. t = 15/(4(1/2sqrt(3) + 3/4))
\ \ \ \ \ \ \= 15/(2sqrt(3) + 3)
\ \ \ \ \ \ \= 10sqrt(3)-15
\ \ \ \ \ \ ~~ 2.3205080 ...
With this value of
A = 1/4sqrt(3) (10sqrt(3)-15)^2 + 1/16(10-3(10sqrt(3)-15)) ^2
\ \ \ \ \ \= 2125/16-75sqrt(3)
\ \ \ \~~ 2.908689 ...
And:
s = 1/4(10-3t)
\ \ = 1/4(10-3(10sqrt(3)-15))
\ \ = 55/4-15/2sqrt(3)
\ \ ~~ 0.75961894 ...
We can visually verify that this corresponds to a maximum by looking at the graph of
graph{1/4sqrt(3) x^2 + 1/16(10-3x) ^2 [-5, 10, -20, 20]}
So minimum area of