i am having serious optimization problems. i don't get it!!! plz help.
a 216-m^2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. what dimensions for the outer rectangle will require the smallest total length of fence? how much fence will be needed?
you are designing a 1000-cm^3 right circular cylindrical can whose manufacture will take waste into account. tehre is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. the total amount of aluminum used up by the can will therefore be
A = 8r^2 + 2(pi)rh
rather than the A = 2(pi)r^2 + 2(pi)rh in Example 4. In example 4 the ratio of h to r for the most economical can was 2 to 1. what is the ratio now?
Considering all rectangles with a given perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
Considering all rectangles with a given perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.
Considering all rectangles with the same area, the square results in the smallest perimeter for a given area.
Considering all rectangles with a given perimeter, one side being another straight boundry, the 3 sided
rectangle enclosing the greatest area has a length to width ratio of 2:1
To solve the first problem, let's break it down step by step:
1. Start by setting up the problem. You have a rectangular pea patch that needs to be enclosed by a fence. The pea patch is 216 square meters, and you need to divide it into two equal parts with a parallel fence.
2. Let's assume the length of the rectangular pea patch is L and the width is W.
3. Since the area of the rectangular pea patch is given as 216 square meters, we can write the equation: L * W = 216.
4. Now we need to find the dimensions of the outer rectangle that will require the smallest total length of fence. To minimize the fence length, we can choose L and W such that they are as close as possible to each other.
5. To divide the rectangular pea patch into two equal parts, we will place the parallel fence along the width W.
6. The length of each divided part will be L/2.
7. The total length of fence required is equal to the perimeter of the outer rectangle plus the length of the parallel fence. Thus, the total length of fence can be calculated as: 2L + W.
8. Substitute the value of W from the area equation (L * W = 216) into the total length of fence equation. We get: Total length of fence = 2L + (216/L).
9. To find the minimum length of fence, we take the derivative of the total length of fence equation with respect to L, set it equal to zero, and solve for L.
10. Differentiate the equation: d(Total length of fence)/dL = 2 - 216/L^2.
11. Set the derivative equal to zero and solve: 2 - 216/L^2 = 0.
12. Simplify the equation: 2L^2 - 216 = 0.
13. Solve for L: L^2 = 108, L = √108, L ≈ 10.39 m.
14. Substitute the value of L back into the area equation to find the width W: 10.39 * W = 216, W ≈ 20.79 m.
15. Therefore, the dimensions of the outer rectangle that require the smallest total length of fence are approximately 10.39 m by 20.79 m. The total length of fence needed can be calculated as: 2L + W.
To solve the second problem:
1. Start by setting up the problem. You are designing a right circular cylindrical can, and you want to find the ratio of h (height) to r (radius) that will minimize the amount of aluminum used.
2. The formula for the total amount of aluminum used is given as: A = 8r^2 + 2πrh.
3. The given information from Example 4 states that the most economical can had a ratio of h to r of 2 to 1.
4. Now, we need to find the new ratio when the top and bottom of the can are cut from squares with side length 2r.
5. To find the new ratio, we need to minimize the amount of aluminum used. We can do this by finding the minimum value of the total amount of aluminum equation A = 8r^2 + 2πrh.
6. Since we want to find the minimum, we can take the derivative of A with respect to r and set it equal to zero.
7. Differentiate the equation A = 8r^2 + 2πrh with respect to r to get dA/dr = 16r + 2πh.
8. Set the derivative equal to zero and solve: 16r + 2πh = 0.
9. Simplify the equation: h = -8r/π.
10. Substitute the ratio given in Example 4 (h to r = 2 to 1) into the equation: 2 = -8r/π.
11. Solve for r: r = -π/4.
12. Since the radius cannot be negative, we discard the negative solution.
13. Therefore, the ratio of h to r for the most economical can when the top and bottom are cut from squares with side length 2r is 2 to 1.