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 Let x be the width of the region and y be the length. Let the perimeter be P and the area A. x y x Draw a diagram and introduce variables for the quantities involved.

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 > –2x > > 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 = = 4x x = 75 A(75) = 300(75)-2(75) 2= 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) = 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.