Question

Question #40c26

Answer 1

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`