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### 5.5 optimization

• 2. Optimization Problems 1. Draw an appropriate figure and label the quantities relevant to the problem. 2. Write a primary equation that relates the given and unknown quantities. 3. If necessary, reduce the primary equation to 1 variable (use a secondary equation if necessary). 4. Determine the desired max/min using the derivative(s). 5. Check solutions with possible values (domain).
• 3. Example: An open box is to be made from a 16” by 30” piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. What size should the squares be to obtain a box with the largest volume? V = l×w×h l =30-2x w =16-2x h = x V = 30-2x ( )× 16-2x ( )× x V = 30x -2x2 ( )× 16-2x ( )= 480x -60x2 -32x2 + 4x3 V = 4x3 -92x2 +480x V ' =12x2 -184x+480 = 0 V ' = 4 3x2 - 46x +120 ( )= 0 V ' = 4 x -12 ( ) 3x-10 ( )= 0 x x 16 30 0 £ x £8 x = 10 3 and x =12 x = 10 3
• 4. Example: An offshore oil well located at a point W that is 5 km from the closest point A on a straight shoreline. Oil is to be piped from W to a shore point B that is 8 km from A by piping it on a straight line under water from W to some shore point P between A and B and then on to B via pipe along the shoreline. If the cost of laying pipe is \$1 million under water and \$½ million over land, where should the point P be located to minimize the coast of laying the pipe? 5 km A P B x 8 – x 8 km W WP = x2 +25 PB =8- x C =1× x2 +25 + 1 2 × 8- x ( ) 0 £ x £8 C' = 1 2 x2 +25 ×2x - 1 2 = 0 x x2 +25 = 1 2 x2 +25 = 2x x2 +25= 4x2 3x2 = 25 x = 5 3 x = - 5 3 C 5 3 æ è ç ö ø ÷ » 8.330127 C(8) » 9.433981 abs max
• 5. Example: You have 200 feet of fencing to enclose two adjacent rectangular corrals. What dimensions should be used to maximize the area? A = 2x× 200- 4x 3 = 400x -8x2 3 P = 4x+3y = 200 y = 200- 4x 3 A' = 3 400-16x ( ) 9 = 0 x = 25 0 £ x £ 200 y = 100 3
• 6. Example: Find the radius and height of the right circular cylinder of largest volume that can be inscribed in a right circular cone with radius 4 inches and height 10 inches. r h V = pr2 h 10-h 10-h r = 10 4 h =10- 5 2 r V = pr2 10- 5 2 r æ è ç ö ø ÷ V =10pr2 - 5p 2 r3 V ' = 20pr - 15p 2 r2 = 0 5pr 4- 3 2 r æ è ç ö ø ÷ = 0 r = 8 3 0 £ r £ 4 r(0) = 0 r 8 3 æ è ç ö ø ÷ » 74.42962 r(4) = 0 abs max h = 10 3
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