A rectangular garden of area 75 sq ft is to be surrounded on 3 sides by a brick wall costing $12 per foot and on one side by a fence costing $7 per foot. Find the dimensions of the garden that the cost of the materials is minimized Read more: A rectangular garden of area 75 sq ft is to be surrounded on 3 sides by a brick wall costing $10 per foot and on one side by a fence costing $5 per foot. Find the dimensions of the garden that the cost of the materials is minimized
Let the dimensions of the garden be
x (fence side) and 75/x (since the area is 75).
So the total cost, C(x)
= 7x+2*12(75/x)+12(7)
= 7x+1800/x+84
To find the minimum cost (at the expense of the shape of the garden)
we calculate
C'(x)=0
7-1800/x^2=0
x=sqrt(1800/7)=16.04'
75/x=4.68'
the answer is wrong
You're right, I mis-calculated the cost of the fourth side (12(7) instead of 12x).
Let the dimensions of the garden be
x (fence side) and 75/x (since the area is 75).
Since there are four sides to the garden:
So the total cost, C(x)
= 7x+2*12(75/x)+12(x)
= 19x+1800/x
To find the minimum cost (at the expense of the shape of the garden)
we calculate
C'(x)=0
19-1800/x^2=0
x=sqrt(1800/19)=9.733' (fence side)
75/x=7.706' (wall side)
C(sqrt(1800/19))=$369.86
A rectangular garden of an area of 208 square feet is to be surrounded on three sides by a brick wall costing $8 per foot and on one side by a fence costing $5 per foot. Find the dimensions of the garden such that the cost of the materials is minimized.
To find the dimensions of the garden that minimize the cost of materials, we need to consider the cost of the brick wall and the fence.
Let's denote the length of the rectangular garden as L and the width as W.
The area of the garden is given as 75 sq ft, so we have LW = 75.
The cost of the brick wall per foot is $12. The garden is surrounded on three sides by the brick wall, so the total cost of the brick wall is 12 * (2L + W) dollars.
The cost of the fence per foot is $7. Since the garden is surrounded on one side by the fence, the total cost of the fence is 7 * L dollars.
To minimize the cost of the materials, we need to minimize the overall cost, which is the sum of the costs of the brick wall and the fence.
Total Cost = Cost of the brick wall + Cost of the fence
= (12 * (2L + W)) + (7 * L)
Now, we can substitute the value of W from the area equation LW = 75:
Total Cost = (12 * (2L + (75 / L))) + (7 * L)
To find the minimum cost, we can take the derivative of the Total Cost with respect to L, set it equal to zero, and solve for L. This will give us the value of L that minimizes the cost.
d(Total Cost) / dL = 0
Differentiating the equation with respect to L:
d(Total Cost) / dL = 24 + (75 / L^2) + 7
Setting it equal to zero:
24 + (75 / L^2) + 7 = 0
Simplifying the equation:
99 + (75 / L^2) = 0
Subtracting 99 from both sides:
(75 / L^2) = -99
Multiplying both sides by L^2:
75 = -99L^2
Dividing both sides by -99:
L^2 = -75 / 99
L^2 ≈ -0.7576
Since the square of a length cannot be negative, there is no real solution for L. This means that there is no minimum cost for this scenario.
It's possible that there may be an error in the provided information or constraints of the problem. Please double-check the values and constraints given.
let the side with the fence be x ft long, then the perpendicular side is 75/x
cost = 7x + 12(75/x)
d(cost)/dx = 7 - 900/x^2 = 0 for a min of cost
7 = 900/x^2
x^2 = 900/7
x = 30/√7 = appr 11.339
so the garden has to be 11.339 by 75/11.339
or 11.34 by 6.61
check: 11.34 x 6.61 = 74.96 , not bad
cost = 7(11.339) + 900/11.339 = 158.75 --- minimum cost
let x = 11 , cost = 7(11) + 900/11 = 158.82 , a higher cost
let x = 11.6, cost = 7(11.5) + 900/11.5 = 158.75 -- just a bit higher
MY answer is correct