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3 years ago

The table below shows the cost of riding in a taxi cab. What is the rate of change?

1 answer:

hjlf3 years ago

3 0

Choose any 2 points from the table:

*10----------------------$25
*15----------------------$35
*20----------------------$45
*25----------------------$55

and find the slope of the line. Supposing we focus on

*15----------------------$35
*20----------------------$45

then:
$45-$35 $10
slope = m = ------------- = ---------- = $2/mile
20-15 5

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I really need help with this Question. The question says ‘ The value of the solid’s surface are is equal to the value of the sol

Sergeu [11.5K]

Units of square inches are never equal to units of cubic inches, so we presume the statement of the problem applies to the numerical values in in² and in³.

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3 years ago

Read 2 more answers

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).