Find the point P in the first quadrant on the curve y=x^-2 such that a rectangle with sides on the coordinate axes and a vertex at P has the smallest possible perimeter.
let the point be P(x,y)
Perimeter = 2x + 2y
= 2x + 2(x^-2)
= 2x + 2/x^2
d(perimeter)/dx = 2 -4/x^3
= 0 for min of perimeter
2 - 4/x^3 = 0
2 = 4/x^3
x^3 = 2
x = 2^(1/3) , (the cube root of 2)
y = 2^(-2/3)
(I really expected the point to be (1,1) but the Calculus shows my intuition was wrong
the perimeter with the above answers is 3.78
had it been (1,1) the perimeter would have been 4)
Try this way:
Perimeter = 2x+2
therefore P = 2x + 2/x^2
d(perimeter)/dx = (2x^3-4)/x^3
solving for x = 2^(1/3)
therefor y was given as y = x^-2
plug x into y to solve for y.
y = 2^(-2/3) or y = (1/4)^(1/3)
check the answer it should be 3.78
Hmm, finding the point P with the smallest possible perimeter, eh? Well, I must admit I'm no mathematician, but I'll give it a shot! Are you ready for some clownish calculations? *Honk honk*
Now, let's analyze the situation. We have a rectangle with sides on the coordinate axes, and one vertex at P on the curve y = x^-2. The perimeter of this rectangle will be given by:
P = 2x + 2y
Since we want to minimize the perimeter, we need to minimize this expression. Let's break it down:
P = 2x + 2y = 2x + 2(x^-2)
Now, we can take the derivative of P with respect to x to find the critical points:
dP/dx = 2 - 4x^-3 = 0
Simplifying:
2 - 4/x^3 = 0
Solving for x, we find:
4/x^3 = 2
x^3 = 2/4
x^3 = 1/2
x = (1/2)^(1/3)
Uh-oh! It seems like we've run into a math problem that involves some complex numbers. I'm afraid I couldn't solve this one with a joke! But worry not, my friend. Proceed with caution, and enlist the aid of a mathematician to help you find that elusive point P. Good luck on your quest for the smallest perimeter!
To find the point P in the first quadrant on the curve y = x^(-2) such that a rectangle with sides on the coordinate axes and a vertex at P has the smallest possible perimeter, we need to minimize the perimeter function.
Let's assume that the side length of the rectangle along the x-axis is x, and the side length of the rectangle along the y-axis is y.
The perimeter of the rectangle is given by P = 2x + 2y.
Since one of the vertices of the rectangle is at P, we can write the equation of the curve as y = x^(-2).
Now, substitute y = x^(-2) into the perimeter equation:
P = 2x + 2(x^(-2))
Simplify the equation:
P = 2x + 2/x^2
To minimize the perimeter, we need to find the critical points of the function. We can do this by finding the derivative of P with respect to x and setting it equal to zero:
dP/dx = 2 - 4/x^3 = 0
Solve for x:
2 - 4/x^3 = 0
Rearrange the equation:
4 = 2x^3
Divide by 2:
2 = x^3
Take the cube root of both sides:
x = ∛2
Since we are looking for a point in the first quadrant, the value of x will be positive. Therefore, x = ∛2.
To find the corresponding value of y, substitute x = ∛2 into the equation of the curve:
y = (∛2)^(-2) = 2^(-2/3)
Simplify:
y = 1/2^(2/3) = 1/∛2 = (∛2) / (2∛2) = (∛2) / 2^(4/3)
Therefore, the point P on the curve y = x^(-2) in the first quadrant, where the rectangle with sides on the coordinate axes and a vertex at P has the smallest possible perimeter, is (∛2, (∛2) / 2^(4/3)).
To find the point P on the curve y = x^(-2) that minimizes the perimeter of the rectangle, we need to set up the problem mathematically.
Let P(x, y) be the point P in the first quadrant on the curve. The perimeter of the rectangle can be calculated by adding the lengths of its four sides.
Since the sides of the rectangle lie on the coordinate axes, the length of the rectangle's sides will be equal to the x-coordinate and y-coordinate of point P, respectively.
Let's say the x-coordinate of point P is x and the y-coordinate of point P is y. Therefore, the length of the sides of the rectangle will be x and y, respectively. The perimeter of the rectangle will be:
P = 2x + 2y
We also know that the point P lies on the curve y = x^(-2). Therefore, y = x^(-2).
Now, we can substitute the value of y in terms of x into the equation for the perimeter:
P = 2x + 2(x^(-2))
Simplifying further:
P = 2x + 2/x^2
To find the minimum perimeter, we need to find the value of x that minimizes the function P.
To do that, we can take the derivative of P with respect to x and set it equal to zero to find any critical points:
dP/dx = 2 - 4/x^3
Setting dP/dx equal to zero:
2 - 4/x^3 = 0
Solving for x:
2 = 4/x^3
x^3 = 4/2
x^3 = 2
x = (2)^(1/3)
So, the x-coordinate of point P is (2)^(1/3).
Now, substitute this x-value back into the equation y = x^(-2) to find the y-coordinate:
y = (2)^(1/3)^(-2)
y = (2)^(2/3)
So, the y-coordinate of point P is (2)^(2/3).
Therefore, the point P on the curve y = x^(-2) that minimizes the perimeter of the rectangle is approximately (1.26, 1.59).