A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $50/ft and on the other three sides by a metal fence costing $40/ft. If the area of the garden is 88 square feet, find the dimensions of the garden that minimize the cost.
length of garden ==== x ft
width of garden ===== y ft
xy = 88
y = 88/x
Cost = 50x +40(2y + x)
= 50x + 40x + 80y
= 90x + 80(88/x) = 90x + 7040/x
dCost/dx = 90 - 7040/x^2 = 0 for a min cost
90 = 7040/x^2
90x^2 = 7040
x^2 = 704/9
x = √704/3 = 8√11/3 ft = appr 8.844 ft
y = 88/8.844 = appr 9.95 ft
To find the dimensions of the garden that minimize the cost, we can use the method of calculus by applying the concept of optimization.
Let's assume the length of the garden to be x feet and the width to be y feet.
Given that the area of the garden is 88 square feet, we have xy = 88.
To minimize the cost, we need to express the total cost C as a function of one variable.
The cost of the brick wall is $50/ft and the cost of the metal fence is $40/ft. The length of the brick wall is y ft, and the length of the metal fence (which consists of three sides) is x ft.
Therefore, the cost function C can be expressed as:
C = 50y + 40(2x + y)
Simplifying this equation, we have:
C = 50y + 80x + 40y
C = 80x + 90y
Now, we substitute the value of y from the area equation xy = 88:
C = 80x + 90(88/x)
To minimize the cost, we need to find the critical points of the cost function C.
To find the critical points, we differentiate C with respect to x and set it equal to zero:
dC/dx = 80 - 7920/x^2 = 0
Solving this equation for x, we get:
x^2 = 99
Taking the positive square root, we obtain:
x = sqrt(99)
Substituting this value of x back into the area equation, we can find the corresponding value of y:
sqrt(99) * y = 88
y = 88 / sqrt(99)
So, the dimensions of the garden that minimize the cost are approximately:
x ≈ 9.9499 feet
y ≈ 8.3176 feet
To solve this problem, we need to minimize the cost of enclosing the garden while meeting the condition of having an area of 88 square feet.
Let's assume the width of the rectangular garden is x feet and the length is y feet.
The area of a rectangle is given by the formula A = length × width. In this case, we know that the area (A) is 88 square feet, so we have the equation:
xy = 88
Now, let's calculate the cost of enclosing the garden.
On one side, the landscape architect wants to use a brick wall costing $50 per foot, so the cost for the length of the wall is 50y.
On the other three sides, the architect wants to use a metal fence costing $40 per foot, so the cost for each side (width and two lengths) is 40(x + y + y).
The total cost (C) is the sum of the cost for the brick wall and the metal fence:
C = 50y + 40(x + y + y)
= 50y + 40(x + 2y)
To minimize the cost, we need to find the values of x and y that minimize C.
Substituting the value of xy from the first equation into the cost equation, we have:
C = 50y + 40(x + 2y)
= 50y + 40x + 80y
= 40x + 130y
Now, we have a cost equation in terms of two variables, x and y. To minimize this equation, we need to calculate its partial derivatives with respect to x and y, and set them equal to zero.
∂C/∂x = 40
∂C/∂y = 130
Setting ∂C/∂x = 0 and ∂C/∂y = 0, we have:
40 = 0 (this is not possible, since 40 is a constant)
130 = 0 (this is not possible, since 130 is a constant)
Since we cannot set ∂C/∂x and ∂C/∂y equal to zero, there are no critical points in the interior of the feasible region. However, since we have a constraint equation, we can find the values of x and y that minimize the cost by substituting the value of xy from the first equation into the cost equation.
Substituting xy = 88 into C = 40x + 130y, we have:
C = 40x + 130y
= 40x + 130(88/x)
= 40x + 11440/x
Now, we have a cost equation in terms of a single variable, x. To minimize this equation, we need to find its derivative with respect to x and set it equal to zero.
dC/dx = 40 - 11440/x²
Setting dC/dx = 0, we have:
40 - 11440/x² = 0
Rearranging the equation, we get:
11440/x² = 40
Cross-multiplying, we have:
11440 = 40x²
Dividing both sides by 40, we have:
286 = x²
Taking the square root of both sides, we have:
x = ±√286
Since we are dealing with a length, the negative solution does not make sense in this context. Therefore, we have:
x = √286
Substituting this value back into the equation xy = 88, we can solve for y:
√286y = 88
Dividing both sides by √286, we have:
y = 88/√286
Therefore, the dimensions of the garden that minimize the cost are x = √286 feet and y = 88/√286 feet.