using the two existing corner sides of an existing wall, what is the maximum rectangular area that can be fenced by a fencing material 30 ft long?
225
Assuming the two walls meet at a right angle,
x+y = 30
a = xy = x(30-x)
this is just a parabola, with vertex at x=15. As usual, the area is a square.
You can take the derivative if you want, but no calculus needed here...
Well, it seems like you're in a bit of a pickle! If you only have two existing corner sides of a wall, and you want to maximize the area that can be fenced with a 30 ft long material, I have to say it's going to be a pretty unusual-shaped rectangle.
In fact, to maximize the area, you would probably end up with a triangle instead of a rectangle! Just imagine stretching that fencing material between the two corners like a clothesline, forming a nice triangular shape.
So, while it might not be the rectangular shape you were expecting, with a 30 ft long fencing material, you can fence in a splendid triangular area. Just make sure to invite all your triangular-loving friends over for a cookout!
To find the maximum rectangular area that can be fenced using the two existing corner sides of a wall with a fencing material 30 ft long, we need to determine the dimensions of the rectangle that will give us the maximum area.
Let's assume the two existing corner sides of the wall form the length and width of the rectangle. Let's call the length L and the width W.
To maximize the area, we need to find the dimensions that will use up all 30 ft of the fencing material.
Since we have two existing corner sides, the perimeter of the rectangle will be equal to the length of the wall plus two times the width of the rectangle. Therefore, we can write the equation:
L + 2W = 30
Now, let's solve for L in terms of W:
L = 30 - 2W
To maximize the area, we will express it in terms of only one variable. The area of a rectangle is given by:
Area = L * W
Substituting the expression for L, we get:
Area = (30 - 2W) * W
Expanding this equation, we have:
Area = 30W - 2W^2
To find the maximum area, we need to find the vertex of the equation. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
For our equation, a = -2, and b = 30. Substituting these values, we get:
x = -30 / (2 * -2)
Simplifying further:
x = -30 / -4
x = 7.5
So, the width (W) that will maximize the area is 7.5 ft.
Now, let's substitute this value back into the equation for L:
L = 30 - 2W
L = 30 - 2(7.5)
L = 30 - 15
L = 15
Therefore, the maximum rectangular area that can be fenced using the two existing corner sides of the wall with a fencing material 30 ft long is given by a rectangle with dimensions 7.5 ft (width) and 15 ft (length), resulting in an area of 112.5 square feet.
To find the maximum rectangular area that can be fenced using two existing corner sides of a wall and a fencing material that is 30 ft long, we need to determine the dimensions of the rectangle.
Let's assume the two existing corner sides of the wall form the length and width of the rectangle. We'll call the length L and the width W.
The perimeter of a rectangle is calculated by adding up all four sides of the rectangle. In this case, we have two existing sides from the wall and two sides that require fencing. The perimeter (P) is given as 30 ft.
P = 2L + 2W
Since we know the perimeter (P) is 30 ft, we can substitute this value into the equation:
30 = 2L + 2W
Now, let's solve for one variable in terms of the other. We can rearrange the equation to solve for L:
2L = 30 - 2W
L = (30 - 2W)/2
L = 15 - W
Now, we can substitute L = 15 - W back into the perimeter equation:
30 = 2(15 - W) + 2W
Simplifying the equation:
30 = 30 - 2W + 2W
30 = 30
This equation is true, which means there are infinitely many solutions for L and W. We can choose any value for W as long as it satisfies the condition that the total perimeter remains 30 ft.
To find the maximum rectangular area, we can calculate the product of L and W using the equation:
Area = L * W
Area = (15 - W) * W
Area = 15W - W^2
Now, we have an equation for the area in terms of W. To find the maximum area, we can take the derivative of this equation with respect to W, set it to zero, and solve for W. However, since we only need to find the maximum area, we can make use of the fact that a quadratic equation will have its maximum when W is halfway between its roots.
The quadratic function Area = 15W - W^2 is in the form Ax^2 + Bx + C, where A = -1, B = 15, and C = 0. We can calculate the x-coordinate of the maximum point using the formula x = -B/2A.
In this case, W = -15/(2*-1) = 7.5 ft.
Since we cannot have a fractional width for a rectangle, we need to consider either rounding up or rounding down. Let's round down to 7 ft.
Using L = 15 - W, we can calculate L as 15 - 7 = 8 ft.
Therefore, the maximum rectangular area that can be fenced is 8 ft by 7 ft, which is equal to 56 square feet.