Question 9:

You need to convert the word problem into a number problem.

Let the two pieces be x and y.

First we know that the string is 6m long therefore

X+Y=6

So Y=6-X …Equation 1

X is shaped into a square so each side is X/4 long and the area of the square is (X/4)^2

Y is shaped into a rectangle with one side b and the other side 3b.

Y is the perimeter so

Y=b+b+3b+3b=8b

So b=y/8

The area of the rectangle is b*3b=3b^2

So in terms of Y the area is 3*(y/8)^2

And substituting in from Equation 1 the area of the rectangle becomes 3*((6-X)/8)^2

We now have the two areas in terms of X

A= (X/4)^2 +3*((6-X)/8)^2

=(x^2)/16 + (3/64)(36-12X+x^2)

The sum of the two areas now has to be maximised.

A maximum occurs when

dA/dX = 0

=2X/16 + (3/64)(0-12+x^2)

=X/8 +6X/64 -36/64

(8X+6X)/64 =36/64

14X=36

X=36/14 =2 4/7

Y=6-X = 3 3/7

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