Optimization ProblemsExample 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 > 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 – 2x2

• 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-2x2

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.