# OPTIMIZATION QUESTION: A wire 60 cm long is to be cut into two pieces. One of the pieces will be bent into the shape of a square and the other into the shape of an equilateral triangle, as shown in the diagram attached. The wire is to be cut in order that the sum of the areas of the square and the triangle is to be a maximum. An equation that can be used to model the sum of the areas is . Determine the boundaries and the corresponding areas. You need not solve further.

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If the 60cm wire is cut into 2 parts labelled x and 60-x, the length x is used to form a square and the length 60-x is used to form an equilateral triangle then:

The side length of the...

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The formula given for the sum of the areas is incorrect.

If the 60cm wire is cut into 2 parts labelled x and 60-x, the length x is used to form a square and the length 60-x is used to form an equilateral triangle then:

The side length of the square is `x/4` and its area is `x^2/16` .

The side length of the triangle is `(60-x)/3` . The area of an equilateral triangle is `A=sqrt(3)/4 s^2` where `s` is the side length. So the area of the triangle would be `sqrt(3)/4 ((60-x)/3)^2=(sqrt(3)(60-x)^2)/36` **The given denominator is 18 which is incorrect**

Thus the sum of the areas is `x^2/16+(sqrt(3)(60-x)^2)/36`

To maximize this sum we take the first derivative and set it equal to zero to find the critical points:

`A(x)=x^2/16+(sqrt(3)(60-x)^2)/36`

`A'(x)=x/8+(1/36)(2sqrt(3))(60-x)(-1)`

`=x/8-(sqrt(3)(60-x))/18`

If `A'(x)=0`

Then `x/8=(sqrt(3)(60-x))/18`

`18x=480sqrt(3)-8sqrt(3)x`

` ` `18x+8sqrt(3)x=480sqrt(3)`

`x=(480sqrt(3))/(18+8sqrt(3))~~26.098` This value is a minimum.

The side length of the square is approximately 6.5245cm with an area of 42.569 square cm.

The side length of the triangle is approximately 11.007cm with an area of 52.458 square cm

To maximize the area we need to look at the endpoints of the interval, namely when x=0 and when x=60. (On a closed interval we are guaranteed an absolute maximum and an absolute minimum -- these can only occur at critical points (the derivative is zero or fails to exist) and at the endpoints of the interval.)

If x=0 the area of the square is 0 and the side length of the triangle is 20 with an area of `100sqrt(3)~~173.205` square cm

If x=60 the side length of the square is 15 with an area of 225 square cm.

Thus to maximize the combined area we form a square of side length 15cm and an area of 225 square cm.

The graph of the total area function:

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