Only #2 pleaseeee help fast

An airplane flying into a headwind travels the 1800-mile flying distance between New York City and Albuquerque, New Mexico in 3

Anton [14]

Answer:

plane speed = 520 mph

wind speed = 20 mph

Step-by-step explanation:

let the speed of the wind be w mph and the speed of the plane be p mph

flying from NY to NM,

time taken = 3 hr 36 min = 3.6 hr

average ground speed = total distance / time taken

= 1800 / 3.6 = 500 mph

since it is a headwind, the plane speed will be slowed down by the head wind, resulting in a lower ground speed.

we can write the following equation:

ground speed = plane speed - wind speed

500 = p - w ------(eq1)

flying from NY to NM (return flight),

time taken = 3 hr 20 min = 3.333 hr

average ground speed = total distance / time taken

= 1800 / 3.333 = 540 mph

on the return trip, the headwind becomes a tailwind, hence the total ground speed would be faster than the plane's air speed , we can write the following equation:

ground speed = plane speed + wind speed

540 = p + w ------(eq2)

with these 2 systems of equations, we can solve for p & w using either substitution of elimination method.

eventually you will end up with

plane speed = 520 mph

wind speed = 20 mph

4 0

3 years ago

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Find the absolute maximum and minimum values of the function f on the set

lara31 [8.8K]

$f_x=4y^2-2xy^2-y^3=y^2(4-2x-y)=0\implies y=0\text{ or }2x+y=4$

$f_y=8xy-2x^2y-3xy^2=xy(8-2x-3y)=0\implies x=0\text{ or }y=0\text{ or }2x+3y=8$

For now, ignore any critical points that occur on the boundary of $\mathcal D$ .

Case 1: If $y=0$ , then either $x=0$ or $2x=8\implies x=4$ , so we have two critical points (0, 0) and (4, 0), both on the boundary.

Case 2: Suppose $2x+y=4$ . Then ...

Case 2a: ... if $x=0$ , then $y=4$ - critical point at (0, 4) on the boundary.

Case 2b: ... if $y=0$ , then $2x=4\implies x=2$ - critical point at (2, 0) on the boundary.

Case 2c: ... if $2x+3y=8$ , then

$\begin{cases}2x+y=4\\2x+3y=8\end{cases}\implies x=1\text{ and }y=2$

- critical point at (1, 2) not on the boundary.

We compute the Hessian matrix of $f(x,y)$ :

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-2y^2&4(2-x)y-3y^2\\4(2-x)y-3y^2&8x-2x^2-6xy\end{bmatrix}$

$\det\mathbf H(1,2)=32>0\text{ and }f_{xx}(1,2)=-8$

which means (1, 2) is a local maximum, where we get $f(1,2)=4$ .

Now we consider the boundary in three parts:

(1) Fix $x=0$ . Then $f(0,y)=0$ . It's possible that 0 is the smallest value of $f(x,y)$ whereever you go.

(2) Fix $y=0$ . Then $f(x,0)=0$ , and we get the same result as in the previous case.

(3) Fix $x+y=6$ . Then $f(x,6-x)=-2x(x-6)^2$ . Denote this function by $F(x)$ . We compute the first derivative and find the critical points:

$F'(x)=-2(x-6)^2-4x(x-6)=-2(x-6)(3x-6)=0\implies x=6\text{ or }x=2$

then compute the second derivative and determine its sign at these points:

$F''(x)=-2(3x-6)-2(x-6)(3)=-12(x-4)\implies\begin{cases}F''(6)=-240\end{cases}$

which indicates that a maximum occurs as $x=6$ and a minimum at $x=2$ . Respectively, we have $6+y=6\implies y=0$ and $2+y=6\implies y=4$ , and $f(6,0)=0$ and $f(2,4)=-64$ .

So the absolute maximum of $f(x,y)$ over $\mathcal D$ is 4 at (1, 2), and the absolute minimum is -64 at (2, 4).