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Andrei [34K]
2 years ago
11

Please help with question 10

Mathematics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

equation of the line is y=x+1

Step-by-step explanation:

the equation of a line has the form y = mx + b

while m is for the slope and b is for the y-intercept.

you can see from the graph that the line intercepts the y-axis at y=1 this is the b.

you can get the slope of the graph by taking two points from the graph lets say (1, 2) and (3, 4) then using this formula:

<h2>m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }</h2>

in our case it will be:

m=\frac{4-2}{3-1} = \frac{2}{2}=1

then just substitute the values for m and b.

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Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra s
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<em>f(x, y)</em> = <em>x</em> ² - 4<em>xy</em> + 5

has critical points where both partial derivatives vanish:

∂<em>f</em>/∂<em>x</em> = 2<em>x</em> - 4<em>y</em> = 0   ==>   <em>x</em> = 2<em>y</em>

∂<em>f</em>/∂<em>y</em> = -4<em>x</em> = 0   ==>   <em>x</em> = 0   ==>   <em>y</em> = 0

The origin does not lie in the region <em>R</em>, so we can ignore this point.

Now check the boundaries:

• <em>x</em> = 1   ==>   <em>f</em> (1, <em>y</em>) = 6 - 4<em>y</em>

Then

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 6 when <em>y</em> = 0

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -2 when <em>y</em> = 2

• <em>x</em> = 4   ==>   <em>f</em> (4, <em>y</em>) = 12 - 16<em>y</em>

Then

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 12 when <em>y</em> = 0

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -4 when <em>y</em> = 2

• <em>y</em> = 0   ==>   <em>f</em> (<em>x</em>, 0) = <em>x</em> ² + 5

Then

max{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 21 when <em>x</em> = 4

min{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 6 when <em>x</em> = 1

• <em>y</em> = 2   ==>   <em>f</em> (<em>x</em>, 2) = <em>x</em> ² - 8<em>x</em> + 5 = (<em>x</em> - 4)² - 11

Then

max{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -2 when <em>x</em> = 1

min{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -11 when <em>x</em> = 4

So to summarize, we found

max{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = 21 at (<em>x</em>, <em>y</em>) = (4, 0)

min{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = -11 at (<em>x</em>, <em>y</em>) = (4, 2)

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