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@ -278,7 +278,7 @@ This value is the lone critical point, and in this case gives the position of th
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### Using `argmax` to identify where a function is maximized
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This value that maximizes a function is sometimes referred to as the *argmax*, or argument which maximizes the function. In `Julia` the `argmax(f,domain)` function is defined to "Return a value $x$ in the domain of $f$ for which $f(x)$ is maximized. If there are multiple maximal values for $f(x)$ then the first one will be found." The domain is some iterable collection. In the mathematical world this would be an interval $[a,b]$, but on the computer it is an approximation, such as is returned by `range` below. Without out having to take a derivative, as above, but sacrificing some accuracy, the task of identifying `x` for where `A` is maximum, could be done with
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This value that maximizes a function is sometimes referred to as the *argmax*, or argument which maximizes the function. In `Julia` the `argmax(f,domain)` function is defined to "Return a value $x$ in the domain of $f$ for which $f(x)$ is maximized. If there are multiple maximal values for $f(x)$ then the first one will be found." The domain is some iterable collection. In the mathematical world this would be an interval $[a,b]$, but on the computer it is an approximation, such as is returned by `range` below. Without having to take a derivative, as above, but sacrificing some accuracy, the task of identifying `x` for where `A` is maximum, could be done with
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```{julia}
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@ -445,10 +445,10 @@ Ethan Hunt, a top secret spy, has a mission to chase a bad guy. Here is what we
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* He can drive a car - usually some concept car by BMW - at $30$ miles per hour, but only on the road.
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For his mission, he needs to go $10$ miles west and $5$ `miles north. He can do this by:
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For his mission, he needs to go $10$ miles west and $5$ miles north. He can do this by:
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* just driving $8.310$ miles west then $5$ miles north, or
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* just driving $10$ miles west then $5$ miles north, or
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* just running the diagonal distance, or
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* driving $0 < x < 10$ miles west, then running on the diagonal
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@ -846,7 +846,7 @@ Is there "symmetry" in the answer between $x$ and $y$?
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yesnoq("no")
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```
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What is you were do do two pens like this back to back, then the answer would involve a rectangle. Is there symmetry in the answer now?
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What is you were to do two pens like this back to back, then the answer would involve a rectangle. Is there symmetry in the answer now?
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```{julia}
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@ -969,7 +969,7 @@ diff(SA, x, x)(xx) > 0
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radioq((
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"This applies the second derivative test to the lone *real* critical point showing there is a local minimum at that point.",
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"This applies the first derivative test to the lone *real* critical point showing there is a local minimum at that point.",
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"This finds the ``4`th derivative of `SA`"
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"This finds the `4`th derivative of `SA`"
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), 1)
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```
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@ -1004,7 +1004,7 @@ numericq(val, 1e-3)
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###### Question Non-Norman windows
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Suppose our new "Norman" window has half circular tops at the top and bottom? If the perimeter is fixed at $20$ and the dimensions of the rectangle are $x$ for the width and $y$ for the height.
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Suppose our new "Norman" window has half circular tops at the top and bottom. If the perimeter is fixed at $20$ and the dimensions of the rectangle are $x$ for the width and $y$ for the height.
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What is the value of $y$ that maximizes the area?
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@ -1091,7 +1091,7 @@ We approach this using `SymPy` and $n=10$
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```{julia}
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#| hold: true
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#| eval: false
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@syms s xs[1:10]
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@syms x xs[1:10]
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s(x) = sum((x-xi)^2 for xi in xs)
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cps = solve(diff(s(x), x), x)
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```
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@ -1274,7 +1274,7 @@ choices = ["exactly four times",
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L"exactly $\pi$ times",
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L"about $2.6$ times as big",
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"about the same"]
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answ = 1
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answ = 3
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radioq(choices, answ)
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```
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@ -1390,7 +1390,7 @@ choices = ["It finds the critical points",
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radioq(choices, 1)
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```
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The polynomial `nu` is what degree in `p`?
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The polynomial `eq` is what degree in `p`?
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```{julia}
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@ -1399,7 +1399,7 @@ The polynomial `nu` is what degree in `p`?
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numericq(4)
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```
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The only positive real solution for $p$ from $nu$ is
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The only positive real solution for $p$ from $eq$ is
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```{julia}
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@ -1420,7 +1420,8 @@ In [Hall](https://www.maa.org/sites/default/files/hall03010308158.pdf) we can fi
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```{julia}
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#| hold: true, echo
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#| hold: true
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#| echo: false
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p = plot(; legend=false, aspect_ratio=:equal, axis=nothing, border=:none)
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b = 2.
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plot!(p, x -> x^2, -b, b)
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@ -1471,7 +1472,7 @@ numericq(a)
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Numerically find the value of $a$ that minimizes the length of the line seqment $PQ$.
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```{juila}
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```{julia}
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#| hold: true
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#| echo: false
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x(a) = -a - 1/(2a)
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Block a user