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Kipish [7]
3 years ago
15

Given constraints: x>=0, y>=0, 2x+2y>=4, x+y<=8 explain the steps for maximizing the objective function P=3x+4y.

Mathematics
2 answers:
kondaur [170]3 years ago
6 0

Answer:

The maximum value of P is 32

Step-by-step explanation:

we have following constraints

x\geq 0 ----> constraint A

y\geq 0 ----> constraint B  

2x+2y\geq 4 ----> constraint C

x+y\leq 8 ----> constraint D

Solve the feasible region by graphing

using a graphing tool

The vertices of the feasible region are

(0,2),(0,8),(8,0),(2,0)

see the attached figure

To find out the maximum value of the objective function P, substitute the value of x and the value of y of each vertex of the feasible region in the objective function P and then compare the results

we have

P=3x+4y

so

For (0,2) ---> P=3(0)+4(2)=8

For (0,8) ---> P=3(0)+4(8)=32

For (8,0) ---> P=3(8)+4(0)=24

For (2,0) ---> P=3(2)+4(0)=6

therefore

The maximum value of P is 32

Blizzard [7]3 years ago
6 0

Answer:

Graph the inequalities given by the set of constraints. Find points where the boundary lines intersect to form a polygon. Substitute the coordinates of each point into the objective function and find the one that results in the largest value.

Step-by-step explanation:   that is the answer on edg.

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3 years ago
Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

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3 years ago
A paper clip is dropped from the top of a 144‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = −
BabaBlast [244]
\bf s(t)=-16t^2+144\implies \stackrel{velocity}{\cfrac{ds}{dt}}=-32t

when does it hit the ground?  check the picture below.

\bf 0=-16t^2+144\implies 16t^2=144\implies t^2=\cfrac{144}{16}\implies t^2=9&#10;\\\\\\&#10;t=\sqrt{9}\implies t=3

at that moment, the velocity is -32(3).

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4 years ago
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