Math 50A February 18, 2011 Quiz #20 Name: David Arnold ...

Math 50A
February 18, 2011
Quiz #20
Name:
David Arnold
Instructions. (10 points) Provide solutions for each of the following questions in the space provided.
(10 pts ) 1. A drilling rig 12 miles offshore is to be connected by a pipe to a refinery onshore, 20 miles down the
coast from the rig. If underwater pipe costs $50,000 per mile and land-based pipe costs $30,000 per mile,
what values of x and y give the least expensive connection? Note: full credit will not be awarded unless
you include a method verifying optimal solution.
Rig
12 mi
x
y
20 − y
Refinery
Solution: Using the Pythagorean theorem:
x 2 = y 2 + 12 2
x 2 = y 2 + 144
x = √ y 2 + 144
The cost for laying the pipe under water is $50,000 per mile. Thus, the cost for laying pipe underwater a
distance of √ y 2 + 144 miles is 50, 000 √ y 2 + 144. Secondly, the cost for laying pipe over land is $30,000
per mile. Thus, the cost for laying pipe 20 − y miles overland is 30, 000(20 − y). Thus, the total cost is:
C(y) = 50000 √ y 2 + 144 + 30000(20 − y)
Note that this is defined and continuous on [0, 20]. The function C is more easily differentiated if we
write this as:
C(y) = 50000(y 2 + 144) 1/2 + 600000 − 30000y
Differentiating:
[ ]
1
C ′ (y) = 50000
2 (y2 + 144) −1/2 (2y) − 30000
C ′ (y) = 50000y(y 2 + 144) −1/2 − 30000
Set this equal to zero to find the critical value.
Cross multiply and divide both sides by 10,000.
0 = 50000y(y 2 + 144) −1/2 − 30000
50000y
30000 =
(y 2 + 144) 1/2
3(y 2 + 144) 1/2 = 5y

Math 50A/Quiz #20 – Page 2 of 2 – Name:
Square both sides:
Solve for y:
9(y 2 + 144) = 25y 2
9y 2 + 1296 = 25y 2
1296 = 16y 2
y 2 = 81
y = 9
Now we move to the Extreme Value Theorem, evaluating C at both endpoints and at the critical value.
Next:
Finally:
Arranging results in a table:
C(0) = 50000 √ 0 2 + 144 + 30000(20 − 0)
= 50000(12) + 600000
= 1200000
C(9) = 50000 √ 9 2 + 144 + 30000(20 − 9)
= 50000 √ 225 + 330000
= 50000(15) + 330000
= 750000 + 330000
= 1080000
C(20) = 50000 √ 20 2 + 144 + 30000(20 − 20)
= 50000 √ 544
= 50000(4 √ 34)
= 200000 √ 34
y C(y)
0 1,200,000
9 1,080,000
20 200, 000 √ 34
Thus, the minimum cost occurs when y = 9. We can find x by substituting y = 9 into:
x 2 = √ y 2 + 144
x 2 = √ 9 2 + 144
x 2 = √ 225
x = 15
Hence, set y = 9 miles down the shore, which forces an underwater length of x = 15 miles. This will
minimize the cost of construction.