Rectangle algebra

Question

Farmer Ed has7500 meters of fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the river, what is the largest area that can be enclosed?

Answer 1

This problem lends itself to calculus

We can express the area of the rectangle in terms of the perimeter.

A=F(x)  we know that a function is at its maximum or minimum when a small change in the function variable results in very little change in the function itself.

 

We state this as dF(x)/dx = 0

or the slope of the function near the value of x where the function is at maximum or minimum is zero.

 

We can plot the parabolic area expression to find this value or we can solve the derivitive of the expression in the perimeter values for the rectangle for that value when the expression yields zero.

 

we have the total length of the fence is 7500 meters.

the perimeter consists of a width and length value

the fence length is equivalent to two times the width plus one times the length

The area is the length times the width

the length can be put in term of the width or vice versa using 7500 = 2*w + 1*L

 

lets solve for L = 7500 - 2*w

 

then the area is A=w(7500-2w) =7500w-2w^2

dA/dw = 7500-4w

 

for review y=mx + b has slope m and dy/dx = m

y= x^2 has slope 2x and dy/dy =2x

y=x^n has dy/dx = nx^(n-1) and the slope can be in general a function of variable.

 

dA/dw is zero when w is 1875

7500-4w = 0 or 4w=7500 and w = 7500/4= 1875

and L is then 3750

7500= L + 2w= L + 2*1875....L= 7500-2*1875= 3750

 

one can plot the equation for the parabola to find the value of the width for which the equation yields the maximum to arrive at this same result.

 

Answer 2

A rectangle has two pairs of sides with equal length.  So there are two sides of length W (for width), and two sides of length L (for...length).  If we add up the total length of the sides, we have

 

W + W + L + L

 

Since the rectangle is missing a side (Farmer Ed does not fence the side along the river), we can remove either l or w from the expression.  It doesn't matter which one, so we'll remove a side of length W.  Now, the total length is

 

W + L  + L = W + 2L

 

We also know that there are 7500 meters of fencing, so

W + 2L = 7500

 

The area of rectangle is

 

A = L * W

 

This equation has two variables, but we can express it using one variable by solving the other equation for either W or L.  Let's solve for W (since this doesn't end up as a fraction):

 

W + 2L = 7500   Subtract 2L from both sides:

W = 7500 - 2L

 

Substitute this value for W into the area formula:

 

A = L * W

A = L * (7500 - 2L)  distribute L:

A = 7500L - 2L²

 

This is a quadratic, and the squared term has a negative coefficient.  This means that the graph of A vs L is a parabola which opens downwards.  So, the graph has a maximum value at the vertex.

 

The x-coordinate of the vertex of a the Ax² + Bx + C is -B/2A.  Notice that x is the independent variable.  In our equation, L is the independent variable, so we can see that B = 7500, and A = -2.

 

So, -B/2A = -7500/2(-2) = 7500/4 = 1875.

 

This means that the area is maximized when L = 1875.

The maximum is area is found when we substitute this value into the equation for area:

 

A = 7500L - L²

A = 7500(1875) - 1875²