let y be the function of x defined implicitly by the equation x^2 -2xy +2y^4=1. find the value of x where y obtains its maximum value.
getting dy/dx implicitly , ....
2x -2x dy/dx - 2y + 8y^3 dy/dx = 0
dy/dx( 8y^3 - 2x) = 2y - 2x
dy/dx = (y-x)/(4y^3 - x)
= 0 for a max/min
so y-x=0 --> y = x
sub back into original
x^2 - 2x^2 + 2x^4 - 1 = 0
2x^4 - x^2 - 1 = 0
(x^2 - 1)(2x^2 + 1) = 0
x^2 = -1/2 ----> no real solution
or
x^2= 1
x = ± 1
if x = 1
1 - 2y + 2y^4 - 1=0
by observation , y = 1 and y = 0
if x = -1
1 + 2y + 2y^4 - 1 = 0
by observation y = -1 and y = 0
when x = 1, y has a max of 1
check and verification, see
http://www.wolframalpha.com/input/?i=plot+x%5E2+-2xy+%2B2y%5E4%3D1
To find the value of x where y obtains its maximum value, we can use the concept of partial derivatives.
Step 1: Differentiate the given equation implicitly with respect to x.
Given equation: x^2 - 2xy + 2y^4 = 1
Differentiating implicitly with respect to x:
2x - 2y(dy/dx) + 8y^3(dy/dx) = 0
Simplifying the equation:
2x + (8y^3 - 2y)(dy/dx) = 0
Step 2: Solve for dy/dx.
Rearrange the equation to isolate dy/dx:
(8y^3 - 2y)(dy/dx) = -2x
dy/dx = -2x / (8y^3 - 2y)
Step 3: Find the critical points.
To find the critical points, set dy/dx equal to zero and solve for x:
-2x / (8y^3 - 2y) = 0
Since the numerator is zero, we have:
-2x = 0
x = 0
Step 4: Determine the maximum value of y.
To determine if the critical point (x = 0) corresponds to a maximum value for y, we can use the second partial derivative test.
Taking the second derivative of the original equation with respect to x:
d^2y / dx^2 = -2(dy/dx)^2 + 8y^3(d^2y/dx^2)
Plugging in dy/dx = -2x / (8y^3 - 2y):
d^2y / dx^2 = -2(-2x / (8y^3 - 2y))^2 + 8y^3(d^2y/dx^2)
Simplifying:
d^2y / dx^2 = -2(4x^2 / (8y^3 - 2y)^2) + 8y^3(d^2y/dx^2)
At x = 0, this becomes:
d^2y / dx^2 = -8y^3(d^2y/dx^2)
Since y^3 is always non-negative, the sign of the second derivative (d^2y/dx^2) determines if it is a maximum or a minimum.
If d^2y/dx^2 < 0, it indicates a maximum value for y.
If d^2y/dx^2 > 0, it indicates a minimum value for y.
If d^2y/dx^2 = 0, the test is inconclusive.
Unfortunately, we cannot determine the sign of d^2y/dx^2 without knowing the second partial derivative of y with respect to x.
To find the value of x where y obtains its maximum value, we need to differentiate the equation with respect to x and solve for dy/dx. Then, we can find the critical points by setting dy/dx equal to zero.
The given equation is x^2 - 2xy + 2y^4 = 1.
Differentiating both sides of the equation with respect to x using implicit differentiation:
d/dx (x^2 - 2xy + 2y^4) = d/dx (1).
Applying the chain rule, we get:
2x - 2y(dy/dx) + 8y^3(dy/dx) = 0.
Rearrange the terms to solve for dy/dx:
(dy/dx)(-2y + 8y^3) = -2x.
dy/dx = -2x / (-2y + 8y^3).
To find the critical points, we set dy/dx equal to zero:
-2x / (-2y + 8y^3) = 0.
Since the numerator is zero when x = 0, we have:
0 / (-2y + 8y^3) = 0.
This implies that the value of x doesn't affect the critical points. Therefore, the critical points occur when the denominator is equal to zero:
-2y + 8y^3 = 0.
Factoring out y, we get:
y(-2 + 8y^2) = 0.
This equation is satisfied when y = 0 or (-2 + 8y^2) = 0.
For y = 0, substitute it back into the original equation to find x:
x^2 - 2(0)x + 2(0)^4 = 1.
x^2 = 1.
So, x can be ±1 when y = 0.
For (-2 + 8y^2) = 0:
8y^2 = 2.
Simplify further:
4y^2 = 1.
y^2 = 1/4.
Taking the square root:
y = ±1/2.
Substituting these values of y back into the original equation, we can find the corresponding values of x.
For y = 1/2:
x^2 - 2(x)(1/2) + 2(1/2)^4 = 1.
x^2 - x + 1/8 = 1.
x^2 - x - 7/8 = 0.
Using the quadratic formula, we can find the solutions for x.
Similarly, for y = -1/2:
x^2 - 2(x)(-1/2) + 2(-1/2)^4 = 1.
x^2 + x + 7/8 = 1.
x^2 + x - 1/8 = 0.
Using the quadratic formula again, we can find the solutions for x.
By calculating the values of x and substituting them back into the equation, we can find the corresponding values of y. Then, we can determine which one yields the maximum value of y.