A gardener wants to fence the largest possible rectangular area using 200 yards of fencing. Find the best length and width of the garden.
I also need help on this question:
Sandra walked to the top of a hill at a speed of 2km/h, turned around and walked down the hill at a rate of 4km/h. The whole trip took 6 hoours. How many kilometers is it to the top of the hill?
I thought it might be 6km but my math teacher checked it and said it wasn't.
hello?Ms. Sue????
A square is a rectangle. Yes. The largest area would be a square measuring 50 yards on each side.
Her average speed is 3 km/hour.
(3 * 6)/2 = ?
W=Width
L=Length
P=Perimeter
A=Area
P=2(W+L)
200=2(W+L) Divide with 2
100=W+L
100-L=W
W=100-L
A=W*L=(100-L)*L
A=100L-L^2
dA/dL=100-2L
Function has minimum or maximum when first derivate=0
100-2L=0
100=2L Divide with 2
50=L
L=50 yd
If second derivate<0 then function has a maximum.
If second derivate>0 then function has a minimum.
Second derivate=-2<0
function has a maximum for L=50 yd
W=100-L
W=100-50=50 yd
A(max)=50*50=2500 yd^2
what do you do with (*)
* is sign for multiply
To find the best length and width of the garden that would maximize the rectangular area, we can use the concept of optimization. Let's break down the problem step by step:
1. Identify the given information:
- The gardener has 200 yards of fencing.
2. Determine what needs to be optimized:
- We need to maximize the rectangular area.
3. Define the variables:
- Let's use "length" (L) and "width" (W) to represent the dimensions of the rectangular garden.
4. Set up the equation:
- The perimeter of a rectangle is given by the formula: perimeter = 2(length + width)
- In this case, the perimeter is equal to the available fencing, which is 200 yards:
200 = 2(L + W)
- Rearranging the equation, we get: L + W = 100
5. Express one variable in terms of the other:
- Let's solve the equation for one variable in terms of the other. We'll solve for L:
L = 100 - W
6. Find the formula for the area of the rectangle:
- The area of a rectangle is given by the formula: area = length × width
- Substituting L = 100 - W into the formula, we get:
Area = (100 - W) × W = 100W - W²
7. Maximize the area:
- To find the best length and width that maximize the area, we need to find the maximum value of the area equation.
- Since the graph of the area equation is a downward-opening parabola, the maximum value occurs at the vertex.
- The vertex of a quadratic equation, in the form of y = ax² + bx + c, can be found using the formula: Vertex = -b/2a
- In our equation, a = -1, b = 100, and c = 0. Plugging in these values, we find:
W_vertex = -100 / (2 * -1) = 50
8. Calculate the length corresponding to the maximum width:
- Using the equation L = 100 - W, we can calculate the length:
L = 100 - 50 = 50
9. Finalize the solution:
- The best length and width of the garden, which maximize the rectangular area, are 50 yards and 50 yards, respectively.
Therefore, the gardener should choose a length of 50 yards and a width of 50 yards to fence the largest possible rectangular area with 200 yards of fencing.
The largest area is a square.
P = 4S
200/4 = S
even if it said rectangular?
so its's 50 yd by 50 yd?