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Question
Math
Posted about 1 month ago

The function ff of the two real variables tt and uu is determined by f(t,u)=etu2/4t+4u11f(t, u)=e^{\wedge} t u^{\wedge} 2 / 4 t+4 u-11
a) Show that (t,u)=(1,2)(t, u)=(1,2) is a stationary point of the function
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Answer from Sia
Posted about 1 month ago
Solution by Steps
step 1
To find the stationary points of the function f(t,u)=etu24t+4u11f(t, u) = \frac{e^t u^2}{4t} + 4u - 11, we need to compute the partial derivatives ft\frac{\partial f}{\partial t} and fu\frac{\partial f}{\partial u} and set them to zero
step 2
Compute ft\frac{\partial f}{\partial t}: ft=t(etu24t+4u11)=u2et(4t1)4t2 \frac{\partial f}{\partial t} = \frac{\partial}{\partial t} \left( \frac{e^t u^2}{4t} + 4u - 11 \right) = \frac{u^2 e^t (4t - 1)}{4t^2} Set ft=0\frac{\partial f}{\partial t} = 0: u2et(4t1)4t2=0 \frac{u^2 e^t (4t - 1)}{4t^2} = 0 This implies 4t1=0t=144t - 1 = 0 \Rightarrow t = \frac{1}{4}
step 3
Compute fu\frac{\partial f}{\partial u}: fu=u(etu24t+4u11)=etu2t+4 \frac{\partial f}{\partial u} = \frac{\partial}{\partial u} \left( \frac{e^t u^2}{4t} + 4u - 11 \right) = \frac{e^t u}{2t} + 4 Set fu=0\frac{\partial f}{\partial u} = 0: etu2t+4=0etu2t=4u=8tet \frac{e^t u}{2t} + 4 = 0 \Rightarrow \frac{e^t u}{2t} = -4 \Rightarrow u = -\frac{8t}{e^t}
step 4
Substitute t=1t = 1 into u=8tetu = -\frac{8t}{e^t}: u=81e1=8e u = -\frac{8 \cdot 1}{e^1} = -\frac{8}{e}
step 5
Verify that (t,u)=(1,2)(t, u) = (1, 2) is a stationary point: ft(1,2)=22e1(411)412=4e34=3e0 \frac{\partial f}{\partial t} \bigg|_{(1, 2)} = \frac{2^2 e^1 (4 \cdot 1 - 1)}{4 \cdot 1^2} = \frac{4e \cdot 3}{4} = 3e \neq 0 fu(1,2)=e1221+4=e+40 \frac{\partial f}{\partial u} \bigg|_{(1, 2)} = \frac{e^1 \cdot 2}{2 \cdot 1} + 4 = e + 4 \neq 0 Since both partial derivatives are not zero, (1,2)(1, 2) is not a stationary point
Answer
(1,2)(1, 2) is not a stationary point of the function f(t,u)=etu24t+4u11f(t, u) = \frac{e^t u^2}{4t} + 4u - 11.
Key Concept
Stationary Points
Explanation
Stationary points are found by setting the partial derivatives of the function to zero. In this case, (1,2)(1, 2) does not satisfy the condition for both partial derivatives being zero.

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