Optimization

Question

What is the minimum number of feet of fencing needed to create a rectangular garden with an area of 576ft?

 

I keep getting 144 as my answer but my friend says it wrong.. Please help!

Answer 1

The rectangle with the maximum area for a given perimeter is the square. This is evident to anyone who has studied analytic geometry! Derivatives are not necessary!

√576 = 24.

The rectangle is 24 ft by 24 ft.

The perimeter is 4(24) or 96 ft.

Answer 2

The area of rectangle is length times width.  The perimeter is twice the sum of length and width.

 

Let length = x

Let width = y

 

xy = 576                             eq1

 

Perimeter = 2x + 2y            eq2

 

Substituting eq1 into eq2 so that eq2 in terms of x.

 

Perimeter = 2x + 2(576x^-1)

Perimeter = 2x + 1152x^-1

 

 

Take the derivative of the perimeter and set it equal to zero.

 

P' = 0

2 - 1152x^-2 = 0

2 - (1152/x²) = 0

 

Multiply both sides of the equation by x² and solve for x.

 

2x² - 1152 = 0

2x² = 1152

 x² = 576

 

   x = 24

 

Evaluate P'(23) and P'(25) to see if the derivative changes from negative to positive.

 

P'(23) = 2 - (1152 / 23²)

         = 2 - 2.17

          = negative value

 

P'(25) = 2 - (1152 / 25²)

         = 2 - 1.84

         = positive value

 

Therefore,

 

x=24 and y=24.

 

Minimum perimeter = 2(24 + 24)

                             = 2(48)

                             = 96 feet

Answer 3

My solution for this problem is 96

 

Here's how I solved it:

We have two equations.

1. XY = 576

2. 2X + 2Y = P (where P is the perimeter)

We want to minimize the perimeter, but first we ought to reduce these equations.

 

From 1. We can rewrite solve for Y = 576/X

Substituting this into 2 gives us:

2. P = 2X + 2(576/X) = 2X +1152/X

To minimize we must take the first derivative of P with respect to X

2'. dP/dX = 2 + 1152(-X^(-2))

 

Setting dP/dX will give us a critical number.

2'. 0 = 2 -1152(-X^-2)

 

Solve for X:

2 = 1152(X^-2)

X^2 = 576

X = 24

 

Plug that into the equation we have for Y

Y = 24.

 

P = 2X+2Y = 96

(Not surprisingly, the minimum perimeter requirement gives us a square.)

 

But lets check that we got it right.

Lets check that we are at a minimum by looking at two nearby values for X, say (23.5 and 24.5)

 

For 23.5, P = 2(23.5) +2(576/23.5) = 47 + 49.02 = 96.02

For 24.5, P = 2(24.5) +2(576/24.5) = 49 + 47.02 = 96.02

 

So we can confidently say that the garden will have a minimum perimeter with 96 feet of fencing. If we keep the garden as a square, of course.

Answer 4

Hi Harb,

 

Let x and y represent length and width respectively.

 

Since the area has to be 576 we have xy = 576 or y = 576/x.

 

The perimeter is given by P = 2x + 2y = 2x + 2(576)/x

 

P' = 2 - 2(576)/x² ,which equals zero when x =24 (-24 is ignored). So y = 24 too.

 

Those values of x and y produce a perimeter of 96. You might want to use the 2nd derivative test (P'') (or equivalent) to convince yourself that this is a minimum.