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OLEGan [10]
3 years ago
9

X≥1 x≤7 y≥-1/3x+6

Mathematics
2 answers:
Goshia [24]3 years ago
7 0

Answer:

Step-by-step explanation:

i am unsure of what you actually need, but i want to help you.

if you were to set the boundaries of the feasibility region it would be as follows:

as points

A(1, 17/3)   B(7, 11/3)   C(7, infinity)   D(1, infinity)

as graphs

y=-1/3x+6 where 1≤x≤7

x=1

x=7

tino4ka555 [31]3 years ago
4 0

Answer:

The feasible region for the system is shown below.

Step-by-step explanation:

The given system of inequality is shown below,

x\geq 1

x\leq 7

y\geq -\frac{1}{3}x+6

The related equation of first two inequalities are x=1 and x=7 respectively. Both are vertical lines solid lines because the points on the lines are included in the solution set.

The first inequality is x\geq 1, so shade the area to the right.

The second inequality is x\leq 7, so shade the area to the left.

The related equation of third inequality is

y=-\frac{1}{3}x+6

Here, the slope of the line is -1/3 and y-intercept is 6.

At x=3,

y=-\frac{1}{3}(3)+6=5

Plot these two points (0,6) and (3,5) on the coordinate plane and connect them by a straight line.

Check the inequality by (0,0).

0\geq -\frac{1}{3}(0)+6

0\geq 6

This statements is false. It means (0,0) is not included in the shaded region of third inequality.

So, shade the area above the related line and related line is a solid line because the sign of inequality is ≥.

Therefore, the common shaded region represents the feasibility region.

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1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co
Olenka [21]

Answer:

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\

Step-by-step explanation:

First, notice that:

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\

g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\

g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\

We proceed to use the chain rule to find g_{r}(\sqrt{2},\frac{\pi}{4}) using the fact that X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta) to find their derivatives:

g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\

Because we know X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta) then:

\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)

We substitute in what we had:

g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)

Now we put in the values r=\sqrt{2}\ and\ \theta=\frac{\pi}{4} in the formula:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})

Because of what we supposed:

g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})

And we operate to discover that:

g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}

g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}

and this will be our answer

3 0
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