Calculus Ship in trouble?

Question

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.

Answer 1

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)

Answer 2

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

^_^

Answer 3

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

Answer 4

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.