Calculus Ship in trouble?

A landscaper plans to enclose a 3000 square foot rectangular region in her botanical garden. She will use shrubs costing $25 per foot along three sides and fencing costing $10 per foot along the fourth side. Determine the minimum total cost for such a project.

Calculus Ship in trouble?
Since we have a rectangular region and a given area, it is best to write down the formula for the area of a rectangle.



A = length x width.



Let's call the length L and the width W. Therefore,



A = LW



However, we're given that A = 3000, so



3000 = LW



On three sides of this rectangle, it costs $25 per square foot, and $10 on the fourth side. Therefore, the best way to represent the cost would be



C = (cost of length side [shrubs]) + (cost of width side [shrubs]) + (cost of length side [shrubs]) + (cos of width side [fencing])



C = 25L + 25W + 25L + 10W

C = 50L + 35W



However, note that 3000 = LW (by the area formula), so we can express W as



W = 3000/L



and then we can replace



C = 50L + 35 [3000 / L], or

C = 50L + 105000/L, which we can merge as a single fraction:



C = [50L^2 + 105000]/L



And we now define this to be our function that we will be minimizing, C(L).



C(L) = [50L^2 + 105000]/L



To solve for the minimum, we have to take the derivative, and then make it zero. We use the quotient rule.



C'(L) = ( [100L][L] - [50L^2 + 105000][1] ) / L^2

C'(L) = ( 100L^2 - 50L^2 - 105000 ) / L^2

C'(L) = ( 50L^2 - 105000) / L^2



And then we make it 0.



0 = ( 50L^2 - 105000) / L^2



We can find the critical points and effectively ignore the L^2.



0 = 50L^2 - 105000

105000 = 50(L^2)

2100 = L^2,



Therefore, L = "plus or minus" sqrt(2100), or

L = +/- sqrt(2100)



Note that L can not be negative; therefore, we get rid of the negative solution, and



L = sqrt(2100)



Thus, the minimum cost would be at L = sqrt(2100). If we wanted to actually _determine_ what the minimum cost as opposed to when it happens, we would just plug this value into our newly created cost function, C(L).



C(L) = [50L^2 + 105000]/L

C(sqrt(2100)) = [ 50 (sqrt(2100))^2 + 105000 ] / sqrt(2100)



= [ 50 (2100) + 105000 ] / sqrt(2100)

= 210000 / sqrt(2100)



We can rationalize the denominator to obtain:



= 210000 sqrt(2100) / 2100

= 100 sqrt(2100)



But to get it in its simplest form, we can reduce the radical to



= 100 (10 sqrt(21))

= 1000 sqrt(21)
Reply:x is "forth" side, so other side is 3000/x



cost is 10x + 25*(2*3000/x+x)

=35x+150,000/x



derivative is:

35 - 150,000/x^2

setting this =0, we get

x = sqrt(150,000/35)



use a calculator to solve it and stick x back into cost.
Reply:Let

x = length of one side

y = length of other side

C = cost



Therefore,

xy = 3000

and

C = 10x + 25y + 10y + 25y



Therefore,

C = 10x + 180000/x



evaluate dC/dx = 0

0 = 10 - 180000/x2



Therefore,

x2 = 18000



Get the square root. Therefore,

x = 60√5 ≈ $ 134.16



The other side is

y = 3000/60√5

y = 10√5 ≈ $ 22.36





Therefore

The minimum cost is

C = 10(60√5) + 25(10√5) + 10(10√5) + 25(10√5)



C = 1200√5 ≈ $ 2,683.28



^_^
Reply:L = length

W = width

C = Cost for project

A = area



A = LW = 3000

L = 3000/W



C = 2*25W + 25L + 10L = 50W + 35L

= 50W + 35(3000/W) = 50W + 105,000/W



dC/cW = 50 - 105,000/w^2 = 0



50 = 105,000/W^2

50W^2 = 105,000

W^2 = 2100

W = sqrt(2100) = 10sqrt(21) = 45.825757 ft



L = 3000/W = 3000/[10sqrt(21)] = 300 / sqrt(21) = 65.465367



C = 50W + 35L =

C = 50(10sqrt(21)) + 35(300/sqrt(21))

C = 500sqrt(21) + 10,500/sqrt(21) = $4,582.57