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4 years ago

What is the maximum value of 2y + x in the feasible region?

Mathematics

1 answer:

emmasim [6.3K]4 years ago

4 0

<h2>Answer:</h2>

The inquiry can't be addressed on the grounds that:

there is no symbol shown among 2y and x,

there is no data on the practical region.

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Andrews [41]

Hello and Good Morning/Afternoon:

Let's take this problem step-by-step:

First off, let's write the line in point-slope form:

$\rm \hookrightarrow (y-y_0)=m(x-x_0)$

(x₀, y₀) any random point on the line
'm' is the value of the slope

Let's calculate the slope:

$\rm \hookrightarrow Slope = \frac{y_2-y_1}{x_2-x_1}$

(x₁, y₁): any random point on the line ⇒ (-2, -6)
(x₂,y₂): any random point on the line that is not (x₁, y₁) ⇒ (2, -3)

$\rm \hookrightarrow slope = \frac{-3--6}{2--2} =\frac{3}{4}$

Now that we found the slope, let's put it into the point-slope form

⇒ we need (x₀, y₀) ⇒ let's use (2,-3)

$(y-(-3))=\frac{3}{4} (x-2)\\y+3=\frac{3}{4} (x-2)$

The equation, however, could also be put into 'slope-intercept form'

⇒ gotten by isolating the 'y' variable to the left

$y = \frac{3}{4}x-\frac{9}{2}$

Answer: $y = \frac{3}{4}x-\frac{9}{2}$ or $y+3=\frac{3}{4} (x-2)$

*Either equations work, put the one that you are the most familiar with

Hope that helps!

#LearnwithBrainly

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Answer:

Step-by-step explanation:

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