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Maxima and minima

1.[2p]

The function f(x,y)=x3-3x+y2-4y has a critical point at (1,2). What is D=fxxfyy-fxy2 there?

CorrectNot quite: 12

2.[2p]

The same f(x,y)=x3-3x+y2-4y has exactly one local minimum. What is its value?

CorrectNot quite: -6

3.[2p]

At the origin, f(x,y)=x2+6xy+4y2 has gradient 𝟎, with fxx=2 and fyy=8. What kind of critical point is it?

Correct
The answer is: A saddle, since $D = 16 - 36 < 0$
The answer is: A saddle, since $D = 16 - 36 < 0$
The answer is: A saddle, since $D = 16 - 36 < 0$

4.[3p]

Use the second order Taylor polynomial of f(x,y)=exsiny at the origin to estimate f(0.1,0.2). Give the estimate to two decimal places.

CorrectNot quite: 0.22

5.[2p]

The Mauna Loa annual means for 2014 to 2018, rounded, were 399, 401, 404, 407 and 409 ppm. With x counting years from 2014, what is the slope of the least squares line, in ppm per year?

CorrectNot quite: 2.6

6.[2p]

For the same data, what is the intercept c of the least squares line y=mx+c? Give it to one decimal place.

CorrectNot quite: 398.8

7.[3p]

Find the absolute maximum of f(x,y)=x2-4x+y2-2y on the square 0≤x≤3, 0≤y≤3.

CorrectNot quite: 3

8.[1p]

Match each function to the behaviour of its critical point at the origin.

  • x4+y4

  • x4-y4

  • x2-y2

  • -x2-y2

  • points

  • tags

  • a saddle confirmed by D=-4

  • a maximum confirmed by D=4

  • second-derivative-test, degenerate

  • 2

  • a saddle the test cannot detect, with D=0

  • a minimum the test cannot detect, with D=0

Show the answer

x4+y4: a minimum the test cannot detect, with D=0 x4-y4: a saddle the test cannot detect, with D=0 x2-y2: a saddle confirmed by D=-4 -x2-y2: a maximum confirmed by D=4

points: 2 tags: second-derivative-test, degenerate

9.[3p]

Which of these statements are true?

Select all that apply

Correct
Correct
Correct
The answer is: The least squares line passes through the point $(\bar{x}, \bar{y})$, The tip of the cone $\sqrt{x^2 + y^2}$ is a critical point even though its gradient does not exist there, An absolute maximum on a closed and bounded region can occur at a boundary point where $\nabla f \ne \mathbf{0}$
The answer is: The least squares line passes through the point $(\bar{x}, \bar{y})$, The tip of the cone $\sqrt{x^2 + y^2}$ is a critical point even though its gradient does not exist there, An absolute maximum on a closed and bounded region can occur at a boundary point where $\nabla f \ne \mathbf{0}$