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 + $by2xy + 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

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

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

Find the first-order partial derivatives fx(x, y) and fy(x, y).

fx(x, y) =

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

fy(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.

$cxy + 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 fxx, fyy, fxy are continuous near P and that the discriminant D(x, y) is D(x, y) = fxx(x, y)fyy(x, y) − fxy2(x, y). Then

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

• If D(a, b) > 0 and fxx(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, fxx, fyy, fxy, and use them to compute D(x, y).

fxx(x, y) = SFgqQUkJGdg=

fyy(x, y) = U2GIbglD1oM=

fxy(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 fxx(x, y) = $c found above, evaluate D($x, $y) and fxx($x, $y).

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

fxx($x, $y) = SFgqQUkJGdg=

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

• If D(a, b) > 0 and fxx(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, fxx, fyy, fxy, and use them to compute D(x, y).

fxx(x, y) = SFgqQUkJGdg=

fyy(x, y) = U2GIbglD1oM=

fxy(x, y) = UYinCekNO0E=

D(x, y) = 5dWJhLEUU4Y=

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

.

Correct.
Incorrect.