Calculus optimization word problem help!?

Question

Sarah's new country home needs electricity. Her new house is in the woods eight miles straight back from the highway(along the road), and twelve miles from the nearest substation (which is next to the highway). It costs $400 per mile to extend existing lines along the highway, $550 per mile along the roads and $850 per mile to run them through the woods. If Sarah wants to minimize the cost of the new lines, find the best path to lay the lines. Be sure to check the cost of the two extremes: following the roads entirely, cutting through the woods entirely. Then use calculus to minimize the cost of following the road for a bit and then cutting through the woods. In your answer be sure to explain how far to follow the highway and what distance it will be through the woods?

Answer 1

Let 12 - x be the distance to be followed along the highway
Then √(64 + x^2) = the distance through the woods.
The cost equation becomes:
C = 400(12 - x) + 850√(64 + x^2)
C = 4800 - 400x + 850√(64 + x^2)
Take the derivative and set it equal to zero:
dC/dx = -400 + 850(1/2)[(64 + x^2)^(-1/2)](2x)
-400 + 850x/√(64 + x^2) = 0
850x/√(64 + x^2) = 400
17x = 8V(64 + x^2)
17x/8 = √(64 + x^2)
289x^2/64 = 64 + x^2
225x^2/64 = 64
x^2 = 64^2/225
x = +/- 64/15 = 4.2667 [take the positive answer]
Mimimum cost:
C = 400(12 - 4.2667) + 850√(64 + 4.2667^2) = $10800
Follow the highway:
12 - x = 12 - 4.2667 = 7.7333miles
Throught the woods:
√(64 + x^2) = √(64 + 4.2667^2) = 9.0667miles
Extreme paths are left to the student.

Answer 2

V = (20 - 2x) (12 - 2x) (x) dV/dx = 4 (3x² - 32x + 60) d²V/dx² = 8(3x - sixteen) 0 = 4 (3x² - 32x + 60) ........ discover table sure factors with dV/dx = 0 x = (2/3) (8 ± ?19) .... discard (2/3) (8 + ?19) because of the fact x can not be larger than 6 x = (2/3) (8 - ?19) is a max because of the fact d²V/dx² < 0 answer: the optimal field is: 4/3 (7+?19) x 4/3 (a million+?19) x (2/3) (8-?19) for a volume of sixty 4/27 (28+19?19) in³ (approx. 262.7 in³)