Question #40c26

1 Answer
Feb 12, 2017

Minimum area of 2.901 occurs when triangle length is 2.321 and square length is 0.760

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 A(t,s), as a function of a single variable and to maximize the total area, A, wrt that variable (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of A wrt the variable.

The total perimeter is that of 3 sides of a triangle and 4 side of a square; we are told that this perimeter is 10

\ \ \ 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, A, as a function of a single variable t, so differentiating wrt t we get:

(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 (dA)/(dt) =0 =>

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 t we have:

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 y=A(t):

graph{1/4sqrt(3) x^2 + 1/16(10-3x) ^2 [-5, 10, -20, 20]}

So minimum area of 2.901 occurs when triangle length is 2.321 and square length is 0.760