In exams: Automatic Generation of Exams in R

coef <- sample(c(2:9, -(2:9)), 3, replace = TRUE)
x <- sample(c(-5:5), 2, replace = TRUE)
H <- matrix(c(2 * coef[1], coef[2], coef[2], 2 * coef[3]),
  nrow = 2, ncol = 2)

ix <- sample(1:4, 1, prob=c(0.35, 0.15, 0.15, 0.35))
ixt <- c("upper left", "upper right", "lower left", "lower right")[ix]
ixn <- c("11", "12", "21", "22")[ix]
sol <- H[ix]

err <- unique(H[-ix])
err <- err[err != sol]
sc <- num_to_schoice(sol, wrong = err, range = -25:25, method = "delta", delta = 1, digits = 0)

plus <- ifelse(coef < 0, "", "+")

Question

Compute the Hessian of the function $$ \begin{aligned} f(x_1, x_2) = r coef[1] x_1^{2} r plus[2] r coef[2] x_1 x_2 r plus[3] r coef[3] x_2^{2} \end{aligned} $$ at $(x_1, x_2) = (r x[1], r x[2])$. What is the value of the r ixt element?

answerlist(sc$questions, markup = "markdown")

Solution

The first-order partial derivatives are $$ \begin{aligned} f'1(x_1, x_2) &= r H[1,1] x_1 r plus[2] r H[1,2] x_2 \ f'_2(x_1, x_2) &= r H[2,1] x_1 r plus[3] r H[2,2] x_2 \end{aligned} $$ and the second-order partial derivatives are $$ \begin{aligned} f''{11}(x_1, x_2) &= r H[1,1]\ f''{12}(x_1, x_2) &= r H[1,2]\ f''{21}(x_1, x_2) &= r H[2,1]\ f''_{22}(x_1, x_2) &= r H[2,2] \end{aligned} $$

Therefore the Hessian is $$ \begin{aligned} f''(x_1, x_2) = r toLatex(H, escape = FALSE) \end{aligned} $$ independent of $x_1$ and $x_2$. Thus, the r ixt element is: $f''_{r ixn}(r x[1], r x[2]) = r sol$.

answerlist(ifelse(sc$solutions, "True", "False"), markup = "markdown")

Meta-information

extype: schoice exsolution: r mchoice2string(sc$solutions) exname: Hessian

exams documentation built on Oct. 17, 2022, 5:10 p.m.