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