Friday

September 19, 2014

September 19, 2014

Posted by **Caeley** on Monday, March 21, 2011 at 10:19pm.

- Algebra 2 -
**Ms. Sue**, Monday, March 21, 2011 at 10:22pmA square will produce the largest area.

- Algebra 2 -
**tchrwill**, Tuesday, March 22, 2011 at 11:39amConsidering all possible rectangles with a given perimeter, the square encloses the greatest area.

Proof:

Consider a square of dimensions "x "by "x", the area of which is x^2.

Adjust the dimensions by adding "a" to one side and subtracting "a" from the other side.

This results in an area of (x + a)(x - a) = x^2 - a^2.

Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Considering all rectangles with a given area, which one has the smallest perimeter?

Letting the dimensions be "x" and "y", the perimeter is P = 2x + 2y and the area is A = xy.

Substituting x = A/y, we get P = 2A/y + 2y = 2Ay^-1 + 2y.

Taking the first derivitive and setting equal to zero, dP/dy = -2Ay^-2 + 2 = 0 or 1 - A/y^2 = 0 or y^2 = A making y = sqrtA.

Substituing, A = xy = x(sqrtA) makes x = sqrtA also making the figure with least perimeter a square.

You might find the following of some interest:

For a given perimeter, the circle encloses the greatest area.

Lets explore a series of regular polygons to see where it takes us.

Let P = the perimeter of the figure.

Let n = the number of sides to the polygon.

Let b = the length of the sides of the polygon.

Let r = the radius of the vertices of the polygon.

Let h = the altitude (apothem) of the triangles formed by the sides and the radii to the vertices of the polygon.

Let A = the enclosed area of the polygon.

Then, s = P/n, r = s/2sin(360/2n) = P/2nsin(360/2n), h = s/2tan(360/2n) = P/2ntan(360/2n) and A = P^2/4ntan(360/2n).

As the number of sides increases, the polygon approaches a circle.

Letting P = 960 for example:

n........4...........6...........8..........16..........32.........64.........90........180........270.......360

A....57,600...66,510...69,529...72,393...73,103...73,279...73,309...73,331...73,335...73,336 sq.ft.

Clearly, the more sides to the polygon, the area approaches the area of the circle with P = 960 where

A = (Pi/4)(960/Pi)^2 = 73,338.

Another way of looking at it is as follows:

Let P = the perimeter of the polygon.

Let n = the number of sides to the polygon.

Let b = the length of the straight sides of the polygon = P/n.

Let h = the altitude (apothem) of the triangles formed by the sides and the radii to the vertices of the polygon.

Let a = the area of one of the n congruent triangles within the polygon.

Let A = the enclosed area of the polygon.

Let r = the radius of the circumscribed circle.

Let s = an increment of arc of the circumscribed circle.

The area a = bh/2 = Ph/2n.

Then, the area A = Phn/2n = Ph/2.

As "n" increases, the polygon approaches the shape of a circle.

Hence, "h" approaches "r" and "b" approaches "s."

Then, the area of the infinitely small incremental sector is a = sr/2.

The area of the full circle is then A = rC/2.

Knowing that C = 2Pi(r), A = r(2Pi(r)/2 = Pi(r^2), the area of the circle as we know it.

Considering all rectangles with a given perimeter, which one encloses the largest area?

The traditional calculus approach would be as follows.

Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x)/2 = Px/2 - x^2.

Taking the first derivitive and setting equal to zero, dA/dx = P/2 - 2x = 0, x becomes P/4.

With x = P/4, all four sides are equal making the rectangle a square.

.....The short side is P/4.

.....The long side is (P - 2(P/4))/2 = P/4.

Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, the square encloses the maximum area.

Considering all possible rectangles with a given perimeter, the square encloses the greatest area.

Proof:

Consider a square of dimensions "x "by "x", the area of which is x^2.

Adjust the dimensions by adding "a" to one side and subtracting "a" from the other side.

This results in an area of (x + a)(x - a) = x^2 - a^2.

Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Considering all rectangles with a given area, which one has the smallest perimeter?

Letting the dimensions be "x" and "y", the perimeter is P = 2x + 2y and the area is A = xy.

Substituting x = A/y, we get P = 2A/y + 2y = 2Ay^-1 + 2y.

Taking the first derivitive and setting equal to zero, dP/dy = -2Ay^-2 + 2 = 0 or 1 - A/y^2 = 0 or y^2 = A making y = sqrtA.

Substituing, A = xy = x(sqrtA) makes x = sqrtA also making the figure with least perimeter a square.

Considering all possible rectangles with a given area, the square has the smallest perimeter

Proof:

Consider a square of dimensions "x "by "x", the area of which is x^2.

Subtract "a" from one side making it (x - a).

Add "b" to the other side making it (x + b).

Since x^2 - ax + b(x - a) = A, "b" must be greater than "a" as "ax" must equal b(x - a).

Therefore, P = 2(x - a) + 2(x + b) = 2x - 2a + 2x + 2b = 4x - 2a + 2b.

With "b" greater than "a", 4x - 2a + 2b results in a greater perimeter P.

Consider the family of rectangles with area 36 sq.units.

The rectangle dimensions and their associated perimeters are:

x......1......2......3......4......6

y.....36....18....12......9......6

P....74....40....30.....26....24 showing that the square has the smallest perimeter.

Considering all right triangles where the sum of the two sides is a constant, which one encloses the greatest area?

Letting the two sides be "a" and "b", a + b = C or a = C - b.

The area is then A = ab/2 = (C - b)b/2 = Cb/2 - b^2/2.

Taking the first derivitive and setting equal to zero, dA/db = C/2 - b = 0 making C = 2b.

Substituting, a + b = 2b makes a = b.

Therefore, the triangle with the greatest area is a right isosceles triangle.

Considering all possible triangles with a given perimeter, the equilateral triangle encloses the greatest area.

Proof:

Let the sides of the equilateral triangle be "a".

Then, the perimeter is P = 3a and the area is A = sqrt[s(s - a)(s - a)(s - a)] (Heron's area formula) where s = 3a/2.

Adjust the length of two of the sides to (a + b) and (a - b).

The semi-perimetr remains at s = 3a/2.

The area becomes A = sqrt[s(s - a)(s - a - b)(s - a + b)] = sqrt[s(s -a)((s - a) - b)((s - a) + b)]

Note that ((s - a) - b)((s - a) + b) = (s - a)^2 - b^2 which is less than (s - a)^2 when the two sides were equal.

Therefore, making any of the three sides smaller and larger than equal results in a smaller enclosed area.

Considering all possible triangles with a given area, the equilateral triangle has the smallest perimeter.

Proof:

Consider equilateral triangle ABC, A at left, B up top and C to the right.

All interior angles are 60 degrees.

Let the sides of the equilateral triangle be "a".

The altitude h = a(sqrt3)/2 and the area is A = a^2(sqrt3)/2.

The starting perimeter is P = 3a.

Assume point B moves to the left, parallel to AC, to where angle BAC = 75 deg, "h" remaining constant.

The area remains constant at A = ah/2 = a^2(sqrt3)/2.

Side AB = h/sin(75) = (a)sqrt3/2sin75.

Side AB shrinks slower than side BC increases due to the relative sines of the respective angles involved.

Thus, while side AB is less than "a" by some increment, side BC is greater than "a" by a larger increment.

Therefore, the perimeter is larger for the same area.

Assume AB perpendicular to AC with angle BAC equal to 90 deg.

Then, BC = sqrt(h^2 + a^2) and the perimeter is P = asqrt3/2 + a + sqrt(h^2 + a^2), clearly greater than 3a.

As point B moves further to the right, the sides AB and BC grow progressively larger making the perimeter greater yet.

Therefore, for a given area, the equilateral triangle has the smallest perimeter.

Considering all rectangles with a given perimeter, one side being provided by a straight given boundry, which one encloses the largest area?

Letting P equal the given perimeter and "x" the short side of the rectangle, we can write for the area A = x(P - 2x) = Px - 2x^2.

Taking the first derivitive and setting equal to zero, dA/dx = P - 4x = 0, x becomes P/4.

With x = P/4, we end up with a rectangle with side ratio of 2:1.

.....The short side is P/4.The traditional calculus approach would be as follows.

.....The long side is (P - 2(P/4)) = P/2.

Therefore, it can be unequivicably stated that of all possible rectangles with a given perimeter, one side being a given external boundry, the rectangle with side ratio of 2:1 encloses the maximum area.

Considering all rectangles with a given area, one side being provided by a straight given boundry, which one has the smallest perimeter?

Letting A = xy equal the given area and "x" the short side of the rectangle, we can write for the perimeter P = 2x + y.

With y = A/x, we derive P = 2x + A/x

Taking the first derivitive and setting equal to zero, dP/dx = 2 - A/x^2 = 0, x = sqrt(A/2)

With the short side x = sqrt(A/2), we end up with y = A/sqrt(A/2) = 2sqrt(A/2) or a rectangle with side ratio of 2:1.

Therefore, it can be unequivicably stated that of all possible rectangles with a given area, one side being a given external boundry, the rectangle with side ratio of 2:1 has the smallest perimeter.

**Answer this Question**

**Related Questions**

Math - A rectangular livestock pen with THREE SIDES of fencing is to be built ...

calculus optimization problem - A farmer has 460 feet of fencing with which to ...

Algebra 2 - One hundred feet of fencing is available to make a rectangular dog ...

Calculus 2 - A farmer wishes to build a fence for 6 adjacent rectangular pens. ...

math - suppose you have 54 feet of fencing to enclose a rectangle dog pen . the ...

geometry - ye has 44 feet of fencing to enclose a rectangular garden. She wants ...

algebra - Ian wants to build a rectangular pen for his animals. One side of the ...

calculus - 1.Ruth has 240 feet of fencing available to enclose a rectangular ...

preculculus - 27. A farmer wishes to enclose a pasture that is bordered on one ...

algebra - A man wants to build a rectangular pen in which to keep his dog. If he...