A dog kennel is to be constructed alongside a house using 60 m of fencing. Find the dimensions that will yield the maximum area.
How do you solve this problem?
I assume you are implying that we don't need a fence along the house.
let the length be y m (the single side)
let the width be x m
so we know that 2x + y = 60
or y = 60 - 2x
Area = xy
= x(60-2x)
= -2x^2 + 60x
At this point I don't if you know Calculus or not.
If you do, then
d(Area)/dx = -4x + 60
= 0 for a max of area
-4x + 60 = 0
x = 15
then y = 30
if you don't take Calculus, we have to complete the square
area = -2(x^2 - 30x)
= -2(x^2 - 30x + 225 - 225)
= -2((x-15)^2 - 225)
= -2(x-15)^2 + 450
so the area is a maximum when x = 15
and then y = 60-30 = 30
by either method,
the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2
To solve this problem, we need to find the dimensions that will yield the maximum area for the dog kennel using 60 m of fencing.
Let's assume the kennel has two equal sides of length x, and two other sides of length y, all connected at right angles.
1. We know that the perimeter of the kennel, which is the sum of all sides, is equal to 60 m:
2x + 2y = 60.
2. We can simplify the equation by dividing both sides by 2:
x + y = 30.
3. We need to express one of the variables in terms of the other variable so that we can substitute it into the area formula. For example, we can express y in terms of x:
y = 30 - x.
4. The area of the kennel is given by the product of the two sides:
A = x * y.
5. Substitute the expression for y into the area formula:
A = x * (30 - x).
6. Expand the equation:
A = 30x - x^2.
7. We want to find the maximum area, so we need to find the value of x that maximizes A. To do this, we can take the derivative of A with respect to x and set it equal to zero:
dA/dx = 0.
8. Take the derivative of A:
dA/dx = 30 - 2x.
9. Set the derivative equal to zero and solve for x:
30 - 2x = 0.
2x = 30.
x = 15.
10. With x = 15, we can substitute it back into the equation for y to find the value of y:
y = 30 - 15 = 15.
11. Therefore, the dimensions that yield the maximum area are x = 15 m and y = 15 m.
So, the dog kennel with sides of length 15 m and 15 m will have the maximum area when using 60 m of fencing.
To solve this problem, we can use calculus to find the dimensions that will yield the maximum area of the dog kennel.
Let's assume the dog kennel is in the shape of a rectangle. Let's denote the length of the rectangle as L and the width as W.
Since the dog kennel is being constructed alongside a house, it will only have three sides enclosed with fencing. Therefore, the perimeter of the dog kennel is given by:
Perimeter = 2L + W
We are given that the total amount of fencing available is 60 m, so we can write the following equation:
2L + W = 60
We want to find the dimensions that will yield the maximum area, so we need to express the area of the rectangle in terms of a single variable.
The area of a rectangle is given by:
Area = Length × Width = L × W
Now, we have two variables (L and W) and one equation (2L + W = 60). We can solve this equation for one variable and substitute it into the area equation.
From the equation 2L + W = 60, we have:
W = 60 - 2L
Substituting this value of W into the area equation, we get:
Area = L × (60 - 2L) = 60L - 2L^2
Now, we have expressed the area of the rectangle as a quadratic equation. To find the dimensions that yield the maximum area, we need to find the vertex of this quadratic equation.
The vertex of a quadratic equation in the form of f(x) = ax^2 + bx + c can be found using the formula:
x = -b/2a
In our case, a = -2, b = 60, and c = 0. So, we have:
L = -60/(2(-2)) = 15
Now, we can substitute this value of L back into the equation 2L + W = 60 to find the value of W:
2(15) + W = 60
30 + W = 60
W = 60 - 30
W = 30
Therefore, the dimensions that will yield the maximum area of the dog kennel are L = 15 m (length) and W = 30 m (width).