Apipe cleaner 20cm long. it is bent into a rectangle. use a quadratic model to calculate the dimensions that give the maximum area.

The dimensions that give the maximum area is 5 cm by 5 cm.

Given:

The perimeter of this rectangle is 20 cm, and formula for perimeter is

P= 2(W+L)

P = 20 cm = 2W + 2L.  

Then W + L = 10 cm,

or W = (10 cm) - L.

The area of the rectangle is A = L·W, and is to be maximized.  

On substituting the values, we get A = L[ (10 cm) - L ], or A = 10L - L²

Note that this is the equation of a parabola that opens down.  With coefficients a = -1, b = 10 and C = 0, we find that the x-coordinate of the vertex (which is the x-coordinate of the maximum as well) is

x = -b / (2a).  Subbing 10 for b and -1 for a, we get:

x = -[10] / [2·(-1)] = 10/2, or 5.

This tells us that one dimension of the rectangle is 5 cm.

Since P = 20 cm = 2L + 2W, and if we let L = 5 cm, we get:

20 cm = 2(5 cm) + 2W, or

10 cm = W + 5 cm, or W = 5 cm.

Therefore, choosing L = 5 cm and W = 5 cm results in a square, which in turn leads to the rectangle having the maximum possible area.

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5 cm by 5 cm

Step-by-step explanation:

The perimeter of this rectangle is 20 cm, and the relevant formula is

P = 20 cm = 2W + 2L.  Then W + L = 10 cm, or W = (10 cm) - L.

The area of the rectangle is A = L·W, and is to be maximized.  Subbing (10 cm) - L for W, we get A = L[ (10 cm) - L ], or A = 10L - L²

Note that this is the equation of a parabola that opens down.  With coefficients a = -1, b = 10 and C = 0, we find that the x-coordinate of the vertex (which is the x-coordinate of the maximum as well) is

x = -b / (2a).  Subbing 10 for b and -1 for a, we get:

x = -[10] / [2·(-1)] = 10/2, or 5.

This tells us that one dimension of the rectangle is 5 cm.

Since P = 20 cm = 2L + 2W, and if we let L = 5 cm, we get:

20 cm = 2(5 cm) + 2W, or

10 cm = W + 5 cm, or W = 5 cm.

Thus, choosing L = 5 cm and W = 5 cm results in a square, which in turn leads to the rectangle having the maximum possible area.