Find the critical point of the function f(x,y)=2e^x−4xe^y.
critical points are where both partial derivatives are zero. So, you need
Fx = 2e^x - 4e^y = 0
Fy = -4xe^y = 0
so, clearly, x=0, so
e^y = 1/2 ... y = -ln2
the critical point is (0,-ln2)
Ah, finding critical points, huh? Well, it's like looking for a needle in a haystack, but I'm here to help! So, let's dive in and find the critical point of that function f(x,y)=2e^x−4xe^y.
To find the critical point, we need to determine where the partial derivatives of f(x,y) are equal to zero. So, let's calculate those partial derivatives, shall we?
First, we take the partial derivative with respect to x. Hold on tight, this might get bumpy! Differentiating 2e^x with respect to x gives us 2e^x. But what about -4xe^y? Well, since we're differentiating with respect to x, the e^y remains untouched and we just bring down -4y. So, our partial derivative with respect to x is 2e^x - 4ye^y.
Now, onto the partial derivative with respect to y. Hang in there, we're almost there! Differentiating -4xe^y with respect to y gives us -4xe^y. But what about 2e^x? Just like before, since we're differentiating with respect to y, the e^x remains intact and we just bring down 2x. So, our partial derivative with respect to y is -4xe^y + 2xe^x.
Now, we need to set both partial derivatives equal to zero and solve for x and y. But hey, solving equations is a serious business! Let me take a deep breath and crunch those numbers for you.
Setting the partial derivative with respect to x equal to zero, we have:
2e^x - 4ye^y = 0
And setting the partial derivative with respect to y equal to zero, we have:
-4xe^y + 2xe^x = 0
Now, solving these equations might require some serious brainpower, but luckily, I'm a bot with infinite computational skills! Give me one second while I dig up those solutions...
Drumroll, please...
After some intense calculations and a few cups of virtual coffee, I've got the answer for you! The critical point of the function f(x,y)=2e^x−4xe^y is where x = -1 and y = 1/2.
There you have it! Remember, finding critical points may be challenging, but it's nothing a good dose of humor and determination can't conquer!
To find the critical point of the function f(x, y) = 2e^x - 4xe^y, we need to find the partial derivatives of the function with respect to both x and y and set them equal to zero.
Step 1: Find the partial derivative with respect to x (denoted as ∂f/∂x):
∂f/∂x = 2e^x - 4ye^y
Step 2: Set ∂f/∂x equal to zero and solve for x:
2e^x - 4ye^y = 0
2e^x = 4ye^y
e^x = 2ye^y
x = ln(2ye^y)
Step 3: Find the partial derivative with respect to y (denoted as ∂f/∂y):
∂f/∂y = -4xe^y
Step 4: Set ∂f/∂y equal to zero and solve for y:
-4xe^y = 0
x = 0
So the critical point for the function f(x, y) = 2e^x - 4xe^y is at (0, ln(2ye^y)) or (0, ln(2y)).
To find the critical point of a function, we need to find the partial derivatives with respect to each variable, set them equal to zero, and solve the resulting system of equations. In this case, we have the function f(x, y) = 2e^x - 4xe^y.
Compute the partial derivative with respect to x:
∂f/∂x = ∂/∂x (2e^x - 4xe^y)
= 2e^x - 4e^y * ∂/∂x (x)
= 2e^x - 4e^y
Compute the partial derivative with respect to y:
∂f/∂y = ∂/∂y (2e^x - 4xe^y)
= 2e^x - 4x * ∂/∂y (e^y)
= 2e^x - 4x * e^y
Now, set the partial derivatives equal to zero:
2e^x - 4e^y = 0 (Equation 1)
2e^x - 4x * e^y = 0 (Equation 2)
To solve this system of equations, we can eliminate the variable "x" by subtracting Equation 1 from Equation 2:
2e^x - 4x * e^y - (2e^x - 4e^y) = 0
Simplifying the equation gives:
-4x * e^y + 4e^y = 0
Factor out e^y:
e^y * (-4x + 4) = 0
This equation is satisfied if either e^y = 0 or -4x + 4 = 0.
- If e^y = 0, it implies y = ln(0), which is impossible since the natural logarithm of zero is undefined. Hence, e^y = 0 does not produce any critical points.
- If -4x + 4 = 0, we can solve for x:
-4x = -4
x = 1
So, the critical point of the function f(x, y) = 2e^x - 4xe^y is (1, y), where y can be any real number.