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natta225 [31]
3 years ago
14

DAY 4 please help i dont understand

Mathematics
2 answers:
pashok25 [27]3 years ago
7 0
9,600
Hope this helps(:
RideAnS [48]3 years ago
5 0
Area of the larger rectangle = 10*16 = 160 cm^2
Area of trapezium = 1/2 * (16+6) * 10 = 110 cm^2

Area of shaded region
= Area of rectangle - Area of trapezium = 160 - 110 = 50cm^2
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Answer:

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Step-by-step explanation:

Since SAM = ZYX

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Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti
mojhsa [17]

D is the set of points,

\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}

Check for critical points:

f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}

Of these 4 points, only 2 belong to D, (-1, 2) and (1, 2), for which we have

\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}

Now look for extrema along the boundary.

  • If x=-2, then

f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2

We have f'(-2,y)>0 for -2 and f'(-2,y) for 2, which indicates a local maximum at y=2 and minima at the endpoints of this boundary. So

\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}

  • If x=2, then

f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2

We have f'(2,y) for 2, so we have extrema at the endpoints of this boundary.

\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}

  • If y=x, then

f(x,x)=9x+1\implies f'(x,x)=9>0

which tells us f is strictly increasing on this boundary, giving the extrema we already know about,

\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}

  • If y=3, then

f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1

We have f'(x,3)>0 for -2 and 1, and f'(x,3) for -1. This indicates a maximum at x=-1 and a minimum at x=1, with

\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}

From this analysis, we find that f attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

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3 years ago
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