What are the dimensions of a rectangle with area 64000 m whose perimeter is as small as possible?

Answer 1

This is a classic Calc I (AP Calc AB) optimization problem. Essentially you have two defining characteristics of this problem:

xy=64000 where x and y are real numbers

f(x,y)=2x+2y is minimized

Although I have written this in terms of a two-variable equation, the first characteristic equation implies:

xy=64000

y=64000/x

So the function f(x,y) can be rewritten as f(x,64000/x)=f(x)=2x+2(64000/x)

f(x)=2x+128000x-1

To finish the problem we must minimize our new one-variable function f(x). To do this, we must first find the first derivative of f(x).

Using the power rule, f(x) is derived as:

f'(x)=(2)(1)x1-1+(128000)(-1)x-1-1

f'(x)=2x0-128000x-2

f'(x)=2-128000x-2

To find the candidate points for maxima and minima, we must find all points where the first-derivative equals zero or is nonexistent. The first derivative equalling zero or being nonexistent implies that the slope of the original function f(x) could change from increasing to decreasing or from decreasing to increasing, which would respectively be maxima or minima.

Since the second part of f'(x) is the term -128000x-2, which is the same as -128000/x2, f'(x) does not exist at x=0, since division by zero is undefined.

By setting f'(x)=0, we can find the x-values that yield a zero-slope on the original function, which means that the slope could be changing at that point from increasing to decreasing or vice-versa.

f'(x)=2-128000/x2

2-128000/x2=0

2=128000/x2

2x2=128000

x2=64000

x=sqrt(64000)

x=-80sqrt(10), x=80sqrt(10)

x~-252.982, x~252.982

You have identified three potential candidates (called critical points) to be minima or maxima on the function f(x). So far:

x=-80sqrt(10)

x=0

x=80sqrt(10)

Consider these values in the context of your problem. x cannot be negative, because it is impossible for the length or width of a rectangle to be negative. Additionally, x cannot be zero, since a rectangle cannot have a length or width of zero. Therefore, you must only investigate the point x=80sqrt(10).

To confirm this is a minimum, which is required by the problem, you must evaluate x-values to both sides on the first derivative, f'(x). If the values of f'(x) change from negative to positive after crossing x=80sqrt(10), then that x-value is a minimum.

Since 80sqrt(10)~252.982, we will use our investigative values on both sides of this to be x=1 and x=300. The investigative values must be between the critical point being investigated and the next critical point. The next critical point less than the one in question is x=0, so our lower investigative value must be greater than 0 (it is). There are no critical points greater than the one we are evaluating, so we can use any upper investigative value. Hint: easy-to-compute values are helpful.

f'(1)=2-128000=-127998

f'(300)=2-(64/45)=(66/65)~1.015

Values greater than our critical point yield positive values on f'(x), while values less than it yield negative values. The first derivative of a function gives the tangent slope at any point on that function, so this means that the slope is changing from decreasing to increasing, which proves that the critical point that was investigated is a minimum on f(x).

Therefore:

x=80sqrt(10)

y=64000/x=64000/80sqrt(10)=800/sqrt(10)

when rationalized, y=x

So, dimensions of 80sqrt(10) m x 800/sqrt(10) m would yield a minimum perimeter to a rectangle of area of 64000 m2

Numerically, these dimensions are approximately 252.982 m x 252.982 m