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andrew-mc [135]

2 years ago

7

I cannot do this question and i am stuck

1 answer:

Mashutka [201]2 years ago

5 0

Answer:

-57

Step-by-step explanation:

$log \frac{ {c}^{7} }{ {a}^{6} {b}^{2} } \\ \\ = log {c}^{7} - log( {a}^{6} {b}^{2} ) \\ \\ = log {c}^{7} - (log {a}^{6} + log{b}^{2} )\\ \\ = 7 \: log {c} - (6 \: log {a}+ 2 \: log{b})\\ \\ = 7 \:( - 9) - [6 \: ( - 5)+ 2 \: (12)] \\ \\ = - 63 - [ - 30 + 24] \\ \\ = - 63 - [- 6] \\ \\ = - 63 + 6 \\ \\ = - 57$

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Find the absolute maximum and minimum values of the following function in the closed region bounded by the triangle with vertice

alukav5142 [94]

$f(x,y)=2x^2-8x+y^2-8y+2$

$f_x(x,y)=4x-8=0\implies x=2$

$f_y(x,y)=2y-8=0\implies y=4$

There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.

Along the boundary $x=0$ , we have

$f(0,y)=y^2-8y+2=(y-4)^2-14$

which attains a maximum value of

$f(0,4)=-14$

Along $y=44$ , we have

$f(x,44)=2x^2-8x+1586=2(x-2)^2+1578$

which attains a maximum of

$f(2,44)=1578$

Along $y=2x$ , we have

$f(x,2x)=6x^2-24x+2=6(x-2)^2-22$

which attains a maximum of

$f(2,4)=-22$

So over the given region, the absolute maximum of $f(x,y)$ is 1578 at (2, 44).

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2 years ago

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Step-by-step explanation: 5/12

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kolbaska11 [484]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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