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How would you go about solving this problem below?
Acme Power has an order for 6800 emergency power plants that can be produced at Acme's plant in Lincoln, Nebraska or at its plant in Tucson, Arizona. If L is the number of units produced in Lincoln and T is the number of units produced in Tucson, find the number of units that should be produced at the Lincoln plant to minimize the total cost if the cost function in dollars is given by C= 0.25L^2 + 0.3T^2 + 16L + 160T.
a. 1460
b. 1960
c. 2040
d. 2460
e. 2640
f. 2960
g. 3240
h. 3840
or is it none of these?
3 Answers
- Anonymous1 decade agoFavorite Answer
Put T = 6800 - L in the equation for C to get a formula in L only.
You then have an expression of the form:
C = pL^2 + qL + r
and can either complete the square or use differentiation to find the minimum cost.
- ironduke8159Lv 71 decade ago
L+T =6800 --> T = 6800-L
C = .25L^2 +.3(6800-L)^2 +16L +160(6800-L)
C= .25L^2 +.3(6800^2 - 2*6800L +L^2)+1088000 -160L
C= .25L^2 +.3*6800)^2 -4080L+.3L^2 +16L+1088000 -160L
C= .55L^2 -4224L + .3(6800)^2 +1088000
dC/dL = 1.1L -4224
Set this equal ) and get L = 3840 <-- answer h
- 037 GLv 61 decade ago
the plan of attack is this:
L+T=6800
so
L= 6800-T and T=6800-L
plug in for every T in the equation with 6800-L then find the derivative and set it = 0 then solve for L when you get:
.25L² + (.3)(6800-L)² + 16L + (160)(6800-L)=C
.25L² + (.3)(6800² - 2(6800)L +L²) + 16L + (160)(6800-L)=C
.25L² + (.3)(6800²) - 0.6(6800)L +(.3)L² + 16L + (160)(6800) -(160L)=C
taking the derivative we get:
.5L +0 - .6(6800) +0.6L+ 16 + 0- 160= 0
1.1L - 4080 - 144=0
1.1L = 4224
L=3840
Answer h fits nicely
