Let x and y be to positive numbers whose product is 500:
(a) Find the maximum sum of x and y
(b) Find the minimum sum of x and y:
2. A cylindrical container can hold a volume of 1 liter. Find the dimensions
of the container that minimizes the surface area.
3. What are the dimensions of the rectangle with largest area that can be
drawn inside a circle of radius R?
4. Find the maximal area of a right triangle with hypotenuse of length L:
#1 I assume that x,y,z are integers, or there is no maximum sum. The minimum sum would be 3∛500 ≈ 23.811
500 = 2^2 5^3 so its factors are
1,2,250
1,4,125
1,5,100
1,10,50
1,20,25
2,2,125
2,5,50
2,10,25
5,5,20
5,10,10
5,25,4
Now pick the largest and smallest sums.
#2 v = πr^2h = 1
so, h = 1/(πr^2)
The surface area is
a = 2πr^2 + 2πrh
= 2πr^2 + 2πr/(πr^2)
= 2πr^2 + 2/r
da/dr = 4πr - 2/r^2
da/dr=0 when r=1/∛(2π)
so, h=∛(4/π)
#3 The diagonal of the rectangle is the diameter of the circle. So,
x^2+y^2 = 4r^2
a = xy = x√(4r^2-x^2)
da/dx = 2(2r^2-x^2)/√(4r^2-x^2)
da/dx=0 when x = r√2
So, as expected, the rectangle is a square with diagonal 2r
#4 as above, it will be half of a square, with legs L/√2.
It's really the same problem, because a right triangle can be inscribed in a circle. Its hypotenuse is the diameter of the circle.
(a) To find the maximum sum of x and y, we need to find the maximum values for both x and y while keeping their product constant at 500. Let's consider the prime factorization of 500: 500 = 2^2 * 5^3.
To maximize the sum, we want to maximize the numbers x and y. Since 5^3 is the largest power of 5 in the prime factorization of 500, we can assign y = 5^3 = 125.
To find x, we divide 500 by y: x = 500/125 = 4.
Therefore, the maximum sum of x and y is 4 + 125 = 129.
(b) To find the minimum sum of x and y, we want to minimize the numbers x and y while keeping their product constant at 500.
Since 2^2 is the smallest power of 2 in the prime factorization of 500, we assign x = 2^2 = 4.
To find y, we divide 500 by x: y = 500/4 = 125.
Therefore, the minimum sum of x and y is 4 + 125 = 129.
As you can see, the maximum and minimum sums are the same, which is quite amusing. It seems like these two numbers have a sense of humor and can't wait to be added up together!
2. To minimize the surface area of a cylindrical container with a volume of 1 liter, we need to find the dimensions that maximize the ratio of height to radius.
Let's assume the height of the cylinder is h and the radius is r. The volume of a cylinder is given by the formula V = πr^2h.
Since the volume is fixed at 1 liter, we can convert it to cubic centimeters (cm^3), which is more appropriate for the dimensions of the cylinder. 1 liter is equal to 1000 cm^3.
So, we have the equation 1000 = πr^2h.
To minimize the surface area A, we need to find the minimum value of A = 2πr^2 + 2πrh. We can express h in terms of r using the volume equation.
Rearranging the volume equation, we get h = 1000 / (πr^2).
Substituting this value of h in the surface area equation, we have A = 2πr^2 + 2πr(1000 / (πr^2)).
Simplifying, A = 2πr^2 + 2000 / r.
To find the minimum value of this function, we can take its derivative with respect to r and set it equal to zero.
Differentiating, we get dA/dr = 4πr - 2000/r^2.
Setting this equation to zero, we have 4πr - 2000/r^2 = 0.
Rearranging, we get 4πr = 2000/r^2.
Simplifying further, we have r^3 = 500 / π.
Taking the cube root of both sides, we get r = (500 / π)^(1/3).
Substituting this value of r back into the equation for h, we have h = 1000 / (π((500 / π)^(1/3))^2).
Simplifying, we get h = (2000^(1/3)) / ((500 / π)^(2/3)).
Therefore, the dimensions that minimize the surface area of the cylindrical container are r = (500 / π)^(1/3) and h = (2000^(1/3)) / ((500 / π)^(2/3)).
3. The dimensions of the rectangle with the largest area that can be drawn inside a circle of radius R are equal to the diameter of the circle.
Since the diameter of a circle is twice the radius, the dimensions of the rectangle are 2R by R.
Now, the area of a rectangle is given by the formula A = length * width. In this case, the length is 2R and the width is R.
Multiplying length and width, we get A = 2R * R = 2R^2.
Therefore, the dimensions of the rectangle that maximize the area are 2R by R, where R is the radius of the circle.
4. To find the maximal area of a right triangle with a hypotenuse of length L, we need to maximize the product of the two legs, which in turn maximizes the area.
Let's assume the two legs of the right triangle are a and b. According to the Pythagorean theorem, we have the equation a^2 + b^2 = L^2.
To maximize the area, we need to maximize the product of a and b, let's call it A = a * b.
To find the maximum value of A, we can use the AM-GM inequality. According to the AM-GM inequality, the arithmetic mean of two numbers is always greater than or equal to their geometric mean.
Applying this to our case, we have (a + b)/2 ≥ √(ab).
Substituting L^2 for a^2 + b^2, we get (L^2/2) ≥ √(ab).
Now, let's solve for ab: ab = (L^2/2)^2 = L^4/4.
Therefore, the maximal area of the right triangle is A = L^4/4.
To solve these optimization problems, we can use calculus techniques. In each problem, we will need to optimize a certain quantity with respect to given constraints.
1. (a) To find the maximum sum of x and y, we need to maximize the sum function f(x, y) = x + y subject to the constraint g(x, y) = xy = 500. We can use the method of Lagrange multipliers to find the maximum. The Lagrangian function is L(x, y, λ) = f(x, y) - λ(g(x, y) - 500).
To maximize L, we set its partial derivatives equal to zero:
∂L/∂x = 1 - λy = 0,
∂L/∂y = 1 - λx = 0,
∂L/∂λ = xy - 500 = 0.
Solving these equations simultaneously, we get x = y = √500, which gives the maximum sum as 2√500.
(b) To find the minimum sum of x and y, we need to minimize the sum function subject to the same constraint. Again, using Lagrange multipliers, we set the partial derivatives of the Lagrangian equal to zero:
∂L/∂x = 1 - λy = 0,
∂L/∂y = 1 - λx = 0,
∂L/∂λ = xy - 500 = 0.
Solving these equations, we get x = y = √500, which gives the minimum sum as 2√500.
2. To minimize the surface area of a cylindrical container with a volume of 1 liter, we need to minimize the surface area formula A = 2πr^2 + 2πrh, subject to the constraint V = πr^2h = 1, where r is the radius and h is the height of the container.
Using the constraint equation, we can express h in terms of r: h = 1/πr^2. Substituting this into the surface area formula, we get A = 2πr^2 + 2πr(1/πr^2) = 2πr^2 + 2/r.
To minimize A, we need to find the critical points by setting the derivative equal to zero:
dA/dr = 4πr - 2/r^2 = 0.
Solving this equation, we get r = √2/2, which gives the minimum surface area. Substituting this value into the expression for h, we find h = 4/√2.
So, the dimensions of the container that minimize the surface area are r = √2/2 and h = 4/√2.
3. The area A of a rectangle inscribed inside a circle of radius R can be expressed as A = 4xy, where x and y are the sides of the rectangle. We need to maximize A subject to the constraint x^2 + y^2 = R^2.
Using the constraint equation, we can express y in terms of x: y = √(R^2 - x^2). Substituting this into the area formula, we get A = 4x√(R^2 - x^2).
To find the maximal area, we need to find the critical points by setting the derivative equal to zero:
dA/dx = 4√(R^2 - x^2) - 4x^2/√(R^2 - x^2) = 0.
Simplifying this equation, we get 4(R^2 - x^2) - 4x^2 = 0, which leads to x^2 = R^2/2 and y^2 = R^2/2.
So, the dimensions of the rectangle with the largest area that can be drawn inside a circle of radius R are x = y = R/√2.
4. To find the maximal area of a right triangle with hypotenuse of length L, let's denote the two legs of the triangle as x and y.
The area A of the triangle can be expressed as A = (1/2) * x * y. We need to maximize A subject to the constraint x^2 + y^2 = L^2.
Using the constraint equation, we can express y in terms of x: y = √(L^2 - x^2). Substituting this into the area formula, we get A = (1/2) * x * √(L^2 - x^2).
To find the maximal area, we need to find the critical points by setting the derivative equal to zero:
dA/dx = (1/2) * √(L^2 - x^2) - (x^2 / √(L^2 - x^2)) = 0.
Simplifying this equation, we get √(L^2 - x^2) - x^2 / √(L^2 - x^2) = 0. Solving this equation is a bit complicated, but we can use calculus techniques like the second derivative test to confirm that the solution is a maximum.
Therefore, the maximal area of a right triangle with hypotenuse of length L is achieved by solving the above equation for x and y, and then calculating A = (1/2) * x * y.