if 60 m of fencing is available for a rectangular garden, one side of which is against a barn, what areb the dimensions of the garden that will give the maximum area?
If the sides are x and y, with x parallel to the barn,
x+2y = 60, so
x = 60-2y
the area is
a = xy = (60-2y)y = 60y-2y^2
da/dy = 60 - 4y
max a when y=15,x=30
Just noticed this was for trig, not calculus. Algebraically, note that a(y) is a parabola, whose vertex occurs where y=60/4 = 15.
To find the dimensions of the garden that will give the maximum area, we can use the concept of optimization.
Let's assume the length of the garden is L and the width is W. Since one side of the garden is against a barn, the perimeter of the garden can be calculated as:
Perimeter = 2L + W
It is given that the total length of the fencing available is 60 m. Therefore,
2L + W = 60
To maximize the area, we need to express the area of the garden in terms of one variable. The area of a rectangle is given by:
Area = Length x Width
Substituting W = 60 - 2L into the equation above, we have:
Area = L(60 - 2L)
Now, we can find the maximum area by finding the critical points of the function. Taking the derivative of the area function with respect to L and setting it equal to zero:
d(Area) / dL = 60 - 4L = 0
Solving this equation, we get:
60 - 4L = 0
-4L = -60
L = 15
Substituting this value of L back into the equation for W:
W = 60 - 2(15) = 30
Therefore, the dimensions of the garden that will give the maximum area are length = 15 m and width = 30 m.
To find the dimensions of the garden that will give the maximum area, we can use the calculus concept of optimization. Let's break down the problem into steps:
Step 1: Identify what needs to be optimized.
In this case, we want to find the dimensions that will maximize the area of the rectangular garden.
Step 2: Define the variables.
Let's assume the width of the garden is 'x' meters. Since one side of the garden is against the barn, the other side will also be 'x' meters. The length of the garden, denoted by 'y' meters, will be perpendicular to the barn.
Step 3: Express the area in terms of the variables.
The area of a rectangle is given by length multiplied by width. In this case, the area of the garden can be expressed as A = x * y.
Step 4: Express any constraints.
We know that the total amount of fencing available is 60 meters. Since one side of the garden is against the barn and does not require fencing, the total length of the three remaining sides should equal 60 meters: 2x + y = 60.
Step 5: Solve the constraint equation for one of the variables.
Let's solve the constraint equation for y in terms of x:
2x + y = 60
y = 60 - 2x
Step 6: Substitute the expression from the constraint equation into the area equation.
A = x * y
A = x * (60 - 2x)
A = 60x - 2x^2
Step 7: Maximize the area.
To find the maximum area, we need to find the critical points of the area equation. We can do this by taking the derivative of the area equation with respect to x and setting it equal to zero:
dA/dx = 60 - 4x = 0
Solving for x:
4x = 60
x = 15
Step 8: Find the corresponding y-value.
Using the constraint equation, substitute the value of x we found into it:
2(15) + y = 60
30 + y = 60
y = 30
Therefore, the dimensions of the garden that will give the maximum area, with 60 meters of fencing available, are a width of 15 meters and a length of 30 meters.