You have 300 meters of fencing with which to build two enclosures. One will be a square, and the other will be a rectangle where the length of the base is exactly twice the length of the height. Give the dimensions of the square and rectangle that minimize the combined area.
Well, let's have some fun with this fencing problem!
To minimize the combined area of both enclosures, we need to find the optimal dimensions for the square and rectangle.
Let's call the side length of the square "s", and the height of the rectangle "h". According to the problem, the base of the rectangle will be twice the height, so the length of the base will be "2h".
Now, let's consider the fencing. For the square, we have four sides that are all equal to "s", so the total fencing used will be 4s.
For the rectangle, we have two sides of length "h" and two sides of length "2h". The total fencing used for the rectangle will be 2h + 2(2h) = 6h.
Since we only have 300 meters of fencing in total, the combined fencing used for both enclosures should be equal to or less than 300 meters. So, we can write the following equation:
4s + 6h ≤ 300
From here, we need to express "h" in terms of "s", so we can substitute it back into the equation to find the optimal dimensions. Well, I'm not a math genius, but I think it's time to bring in the reinforcements...the quadratic formula!
Let's use the quadratic formula to solve for "h" in terms of "s":
h = (-6s + sqrt(36s^2 - 2400s)) / 12
Simplifying this equation is no laughing matter, but we can see that the discriminant (the expression inside the square root) must be non-negative for real solutions to exist. In this case, the discriminant, 36s^2 - 2400s, should be greater than or equal to zero.
By solving the quadratic inequality, we find that s should be between approximately 0 and 66.67.
Ultimately, to minimize the combined area, you'll need to do some calculations to find the optimal dimensions for the square and rectangle. Sorry if I clown around a lot, but math is serious business! Good luck with your calculations!
To minimize the combined area, we need to find the dimensions of both the square and the rectangle that make the areas as small as possible.
Let's start by finding the dimensions of the square.
1. Let x be the length of one side of the square.
2. Since the square has four equal sides, the total length of fencing used for the square will be 4x.
3. We are given that the total length of fencing available is 300 meters, so we have the equation 4x = 300.
4. Solve for x: x = 300/4 = 75.
5. Therefore, the length of one side of the square is 75 meters.
Now let's find the dimensions of the rectangle.
1. Let y be the height of the rectangle.
2. According to the question, the length of the base is exactly twice the length of the height, so the length of the base is 2y.
3. The perimeter of the rectangle will be 2y + 2(2y) = 6y.
4. We are given that the total length of fencing available is 300 meters, so we have the equation 6y = 300.
5. Solve for y: y = 300/6 = 50.
6. Therefore, the height of the rectangle is 50 meters, and the base is 2 times the height, which is 2 * 50 = 100 meters.
In summary, to minimize the combined area, the square should have dimensions of 75 meters by 75 meters, and the rectangle should have dimensions of 50 meters by 100 meters.
To answer this question, we need to find the dimensions of the square and rectangle that will minimize the combined area. Let's break down the problem step by step:
1. Define the variables:
Let's assume the side length of the square is 's'.
Let's assume the height of the rectangle is 'h', then the length will be '2h' because it is twice the height.
2. Calculate the perimeter constraints:
We have 300 meters of fencing, so the perimeter of the square will be 4s, and the perimeter of the rectangle will be 2h + 4(2h) = 10h.
3. Express the constraint using the perimeter equation:
Since the total length of the fencing is 300m, we can equate the perimeters of the square and the rectangle to 300, which gives us the following equation:
4s + 10h = 300
4. Solve for one variable in terms of the other:
We can isolate 's' in terms of 'h' by rearranging the equation:
4s = 300 - 10h
s = (300 - 10h)/4
5. Express the combined area in terms of one variable:
The area of the square is s^2, and the area of the rectangle is (2h) * h = 2h^2. Therefore, the combined area is:
A = s^2 + 2h^2
6. Substitute the expression for 's' from step 4 into the combined area equation:
A = ((300 - 10h)/4)^2 + 2h^2
7. Simplify the equation:
Expand and simplify the squared term, and combine like terms:
A = (90000 - 6000h + 100h^2)/16 + 2h^2
A = (100h^2 - 6000h + 90000 + 32h^2)/16
A = (132h^2 - 6000h + 90000)/16
8. Find the derivative:
To find the minimum value of the area, we need to find the critical points. Take the derivative of the area equation with respect to 'h':
dA/dh = (264h - 6000)/16
9. Set the derivative equal to zero and solve for 'h':
(264h - 6000)/16 = 0
264h - 6000 = 0
264h = 6000
h = 6000/264
h ≈ 22.73
10. Substitute the value of 'h' back into the perimeter equation to find 's':
4s + 10h = 300
4s + 10(22.73) = 300
4s + 227.3 = 300
4s = 72.7
s = 72.7/4
s ≈ 18.18
So, the dimensions that minimize the combined area are approximately 18.18 meters for the side of the square and approximately 22.73 meters for the height of the rectangle.