The function $f$ of the two real variables $t$ and $u$ is determined by $f(t, u)=e^{\wedge} t u^{\wedge} 2 / 4 t+4 u-11$
a) Show that $(t, u)=(1,2)$ is a stationary point of the function

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the stationary points of the function $f(t, u) = \frac{e^t u^2}{4t} + 4u - 11$, we need to compute the partial derivatives $\frac{\partial f}{\partial t}$ and $\frac{\partial f}{\partial u}$ and set them to zero. ‖ ‖ step 2 ⋮ Compute $\frac{\partial f}{\partial t}$: $$ \frac{\partial f}{\partial t} = \frac{\partial}{\partial t} \left( \frac{e^t u^2}{4t} + 4u - 11 \right) = \frac{u^2 e^t (4t - 1)}{4t^2} $$ Set $\frac{\partial f}{\partial t} = 0$: $$ \frac{u^2 e^t (4t - 1)}{4t^2} = 0 $$ This implies $4t - 1 = 0 \Rightarrow t = \frac{1}{4}$ ‖ ‖ step 3 ⋮ Compute $\frac{\partial f}{\partial u}$: $$ \frac{\partial f}{\partial u} = \frac{\partial}{\partial u} \left( \frac{e^t u^2}{4t} + 4u - 11 \right) = \frac{e^t u}{2t} + 4 $$ Set $\frac{\partial f}{\partial u} = 0$: $$ \frac{e^t u}{2t} + 4 = 0 \Rightarrow \frac{e^t u}{2t} = -4 \Rightarrow u = -\frac{8t}{e^t} $$ ‖ ‖ step 4 ⋮ Substitute $t = 1$ into $u = -\frac{8t}{e^t}$: $$ u = -\frac{8 \cdot 1}{e^1} = -\frac{8}{e} $$ ‖ ‖ step 5 ⋮ Verify that $(t, u) = (1, 2)$ is a stationary point: $$ \frac{\partial f}{\partial t} \bigg|_{(1, 2)} = \frac{2^2 e^1 (4 \cdot 1 - 1)}{4 \cdot 1^2} = \frac{4e \cdot 3}{4} = 3e \neq 0 $$ $$ \frac{\partial f}{\partial u} \bigg|_{(1, 2)} = \frac{e^1 \cdot 2}{2 \cdot 1} + 4 = e + 4 \neq 0 $$ Since both partial derivatives are not zero, $(1, 2)$ is not a stationary point. ‖ ∻Answer∻ ⚹ $(1, 2)$ is not a stationary point of the function $f(t, u) = \frac{e^t u^2}{4t} + 4u - 11$. ⚹ ∻Key Concept∻ ⚹Stationary Points⚹ ∻Explanation∻ ⚹Stationary points are found by setting the partial derivatives of the function to zero. In this case, $(1, 2)$ does not satisfy the condition for both partial derivatives being zero.⚹◊What is the definition of a stationary point in calculus?⍭ Generate me a similar question◊

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