Chapter 1. calc_tutorial_14_7_007

1.1 Problem Statement

{6,7,8}

{2,5}

2*$a

2*$b

-(round(1/($c*$d-1),3))

$d*$y

$c*$d-1

Find the critical point of the function. Then use the Second Derivative Test to determine whether it is a minimum, maximum, or saddle point.

f(x, y) = $ax² + $by² − xy + x

1.2 Step 1

Question Sequence

Question 1.1

Recall the definition of a critical point.

A point P = (a, b) in the domain of f(x, y) is a critical point if

f_x(a, b) = 1Wh3cvJ2xF4= or f_x(a, b) does not exist and if

f_y(a, b) = 1Wh3cvJ2xF4= or f_y(a, b) does not exist.

Find the first-order partial derivatives f_x(x, y) and f_y(x, y).

f_x(x, y) =

bTLZyugzFQykgeZTQ8HOG48OiRP19CpVO0l6wew3hp43y1iGO5UJF1+WRQCNSYPyIDDJfQeOpzJMk0QmxiwUVhIIskoy+E6luThPa+bF5dg+NM7N0O75P+h2qxx5yqI4Z+NLtnmd4/vnmyxoJnvZoabhfXtDhmvQpy6wIjACsm4czfEkvQ9B2HU7Ymq7XlksN6y1f0GX88uZfNG8d20/elUbmo9CxAQTBTXcp2s1iArGB4yjTDjw30L1mIPp3bMmONdlHhDy9ktbNyqtdGd/o2BCeTI5ms9fZMB+Gp+gAMweQlTITgsIV2bjremeTb72a9WyoVup/nZYVQU2apGHLnsdGmXq7BThkoEBehBXC8EUNOtZBR1pwt1/mVosLbbA+BvxrXvGHEs9pfp3v4VzvTcA0U7UWjVFhBCkXg4YXQ3WU8fgZuH6wN8A/dMx3mNpTfSv+e8OdQkWRFjjVATrBL78d4avE5BqgGGRSePRqPZ6aHTHeSQhe+a9Wf3h07r3lrq5/oN3XuuBSQrF0otF8HpLtMUy7I0uSmmZZaeFd1Xi7LHwuG6JpF452RxxEChzAoXKskfZXqrXzoqR+P0hAlsUPYpDEt3DoExCc5kPP2FOW/YVaMugJ3N6QS8rJXTBQUj1oWxHYF8lQk+EF1iKipj4Ez8whqPN6ePe5tDnMJzGjM2AP7Sc3bk8lm82QB3vMaxTphtzAWsWNw1l2hm1CFqA5smvqqdq9Uvigz0plbrACaPITL10nxy7pbTNWWI4kMBoZahFK83Zipp8bnUYQiOT6UcV9kywZ02CmdZQAOaH2eh8BI+SWgqdOHjWLvSNOSVFBLJkYndDyM1m1QA3WIYgiMh4UIMzbLAR4Gzqs1QR3fLATme9qBeDrX5jsZne5vLznitv4kUKIQG29DBrv1WpqS7+Nahg4PNXQNGE9kz36HfP/aYxwRt4Xj/nyo6gUlL5LTVlL1c=

f_y(x, y) =

dDJJI1L0WF+vzb17yAEXyj6r8KRH77SIh4QpRSS6XEG6im2b/4eFb1gLRVPSwxyYTc1H8XPIT9yG0WlBdAHmI874Brk7L901cyKNjz4POZ4yjSeSumeqUXJca9pWVU0117hVMQjW5A431slH26qo1z4AwnH6ongWPA+Qmt7lA+aw2j+Or4K3RUhFocS404r6i1qLHabj+Fj0GPCeg5PAUEGDp6BDTNzoIJ+yscHMcgjICrUWpOLcvMsEZ2Dkpw7l4n0vezt83pdBYwZPJ0dTu++YHDLAgL4+Dd7UxfuaHi0nKWMzHjRP0C7jczZgTGdUrPZR0YY2Y7RVeTUSoLzenZPz9nqOdswY9L1KOdMn5QDg7HV++I0ivZ1foJraw0nddXzuwGg26xD869VGNinuNnnuMLE37BbiDS8mauP3GImcvDQWBEIlStpIOrDFD9qq9lqmGtf5zRBPhgLUe4Pm/xi1+QYzrp/bk2yQ/YjkzkdX/Jvs//CGeIUXsQC20zVg6snC3RRI8cwMeq2g8d6FEGUPnyLMI75PMaYJJcZKZPhPahYxxZiMhKNTRnmohjlolhfvm3d72wDmEKq0CG8hoO/qDuTJi28LJFjFR1wfqqCJ+3tGV1A++cD1A8tQ8j64yax3EAdtplroaQaPG3GVlzb8vALNuHixONIXtkT+6hOqJgSDCKLKtICOuVFuA0JcIv5qURieDRMkrkvqlhOgnXHLZtcnBbnZa+8IV7CZPIBwvQSI1Otm+NiNQh+zf10F8b5MlxyeP67Xv1eOOCXj+gPauDyIi5Y8g9J7iPaXjYG+M1+AEU8NKwG9ws5kpMjE6HUB0xywKoCqACsCb1WmtRCBErWCKsosK+/Nej7UDQvFgKmRjiUSRFyhO0hNVQhge4iqptVIsj60xmyCB4E3+56cbAI2YBlEMl0bX9fZnI0zo6jYFvRIwzSXUIk10YcqwbguakaIJtU=

Correct.

Incorrect.

1.3 Step 2

Question Sequence

Question 1.2

Set the partial derivatives equal to zero and solve the system of equations for the critical points. Since the partial derivatives are polynomials, they exist everywhere. Hence the solutions to the system of equations will yield all the possible critical points.

$cx − y + 1 = 0

$dy − x = 0

With coordinates rounded to three decimal places, the critical point is

(x, y) = (i0sYGeK9R9c=, K72fwlfIjEI=)

Correct.

Incorrect.

1.4 Step 2

Question Sequence

Question 1.3

Since ($x, $y) is the only solution to the system of equations, it is the only critical point. Recall the Second Derivative Test for classifying the type of critical point. Let P = (a, b) be a critical point of f(x, y). Assume f_xx, f_yy, f_xy are continuous near P and that the discriminant D(x, y) is D(x, y) = f_xx(x, y)f_yy(x, y) − f_xy²(x, y). Then

• If D(a, b) > 0 and f_xx(a, b) > 0, then f(a, b) is bWpqbTfquXUibB+ftka25pgc7n5p7G5l91SY7tSBcRXyKzHtOOjrwLTz5FX95gJqUjsz0bBAF5FkJA5Aq5Likg==.

• If D(a, b) > 0 and f_xx(a, b) < 0, then f(a, b) is w1fsLWD2WqFjkuvMUzyEfyWHn7AOkZ+rAH7R7ZJ/Z5mjMgJzn/sDnr6wD3kU5l8W4Wr+sVbYBXQwfXM17wum5g==.

• If D(a, b) < 0, then f(a, b) is asy9XFoCIcLtjk9du/z2Ndr2EdcPzt0i7kQk3/uBhTUC/UY8ppL//D2+AUJnwEqCbRbFhXKehn1wXpepbQGRHw==.

• If D(a, b) = 0, then the test is Os0UJxCC+g+g6V9/P0dpi9HfIDWICnVd9UNp4dJN+x4y4S3X6L63LvwJUCY=.

Find the second-order partial derivatives, f_xx, f_yy, f_xy, and use them to compute D(x, y).

f_xx(x, y) = SFgqQUkJGdg=

f_yy(x, y) = U2GIbglD1oM=

f_xy(x, y) = UYinCekNO0E=

D(x, y) = 5dWJhLEUU4Y=

Correct.

Incorrect.

1.5 Step 2

Question Sequence

Question 1.4

Use the Second Derivative Test to classify the critical point ($x, $y). Using D(x, y) = $e and f_xx(x, y) = $c found above, evaluate D($x, $y) and f_xx($x, $y).

D($x, $y) = 5dWJhLEUU4Y=

f_xx($x, $y) = SFgqQUkJGdg=

• If D(a, b) > 0 and f_xx(a, b) > 0, then f(a, b) is bWpqbTfquXUibB+ftka25pgc7n5p7G5l91SY7tSBcRXyKzHtOOjrwLTz5FX95gJqUjsz0bBAF5FkJA5Aq5Likg==.

• If D(a, b) > 0 and f_xx(a, b) < 0, then f(a, b) is w1fsLWD2WqFjkuvMUzyEfyWHn7AOkZ+rAH7R7ZJ/Z5mjMgJzn/sDnr6wD3kU5l8W4Wr+sVbYBXQwfXM17wum5g==.

• If D(a, b) < 0, then f(a, b) is asy9XFoCIcLtjk9du/z2Ndr2EdcPzt0i7kQk3/uBhTUC/UY8ppL//D2+AUJnwEqCbRbFhXKehn1wXpepbQGRHw==.

• If D(a, b) = 0, then the test is Os0UJxCC+g+g6V9/P0dpi9HfIDWICnVd9UNp4dJN+x4y4S3X6L63LvwJUCY=.

Find the second-order partial derivatives, f_xx, f_yy, f_xy, and use them to compute D(x, y).

f_xx(x, y) = SFgqQUkJGdg=

f_yy(x, y) = U2GIbglD1oM=

f_xy(x, y) = UYinCekNO0E=

D(x, y) = 5dWJhLEUU4Y=

Since D($x, $y) cMlp0KGJU5xcQ8H72EPbvw== 0 and f_xx($x, $y) cMlp0KGJU5xcQ8H72EPbvw== 0, the point ($x, $y) is Jc3hN4lFiBNsjMr7Pf9NAQj1x83kiBYe/fhScjOdcfBytBSlckTbO4d6rC5SmGdx/H/bGrrJRwIN/BT1f6gK+g==.

Correct.

Incorrect.