Optimization Problems Example 1: A rancher has 300 yards of fencing material and wants to use it to enclose a rectangular region. Suppose the above region is bordered by a river so that fencing is only needed on three sides. What dimensions would give a region of maximum area?

Solution Draw a diagram and introduce variables for the quantities involved.

Let x be the width of the region and y be the length. Let the perimeter be P and the area A.

x x y

Identify the quantity to be maximized or minimized. Write an expression for this quantity.

Maximize the area. A = xy

Identify constraints and write an expression for the constraints.

The constraints are that the perimeter can only be 300 yards. (that is all the fencing that we have) Therefore: P = 2x +y 300 = 2x + y (since the perimeter is 300)

Rewrite the Constraint formula • 300 = 2x + y • y = 300 –2x • Also x and y must be greater than zero.

• x > 0 , y >0 • Since y = 300 –2x, and y > 0 • 300 –2x > 0 • 300 > 2x • x < 150 • So 0 < x < 150 • These are the endpoints that we must check later.

Write the expression to be maximized as a FUNCTION.

• Use the constraint formula to substitute into the maximized expression.

• A = x y • Substitute y = 300 – 2x into A = xy • A = x (300 –2x)

Find the critical points for this function.

• A(x) = x (300 –2x) • A(x) = 300x – 2x 2 • A’(x) = 300 – 4x • Set A’(x) = 0 to find critical values.

• 300 – 4x = 0 • 300 = 4x • x = 75 • A(75) = 300(75)-2(75) 2= 11250 • The critical point is (75,11250)

Graph of A(x)=300x-2x 2

Determine if this is a maximum or a minimum.

• Use the second derivative test • A’’(x) = -4 • Since A’’(x) < 0 for all x values, (75, 11250) is a maximum.

Check the Endpoints • A(0)= 0 • A(150) = 0 • A(75) = 11250 • Therefore when x = 75 there is an absolute maximum for the given interval.

State the Solution.

• Substitute x = 75 into y = 300 –2x • y = 300 – 2(75) • y = 150 • Therefore the dimensions that will maximize the area of the region is a width of 75 yards and a length of 150 yards.