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

So minimum area of