Convert the following equation into simultaneous equation and solve
1 answer:
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Answer:
Step-by-step explanation:
for the perimeter we just need the distances from point to point
P1(1,1) (x1,y1)
P2(6,2) (x2,y2)
P3(5,4) (x3,y3)
P4(2,5) (x4,y4)
from P1 to P2 is
dist = sqrt [ (x2-x1)^2 +(y2-y1)^2 ]
dist =
from P2 to P3 is
dist = sqrt [ (x3-x2)^2 +(y3-y2)^2 ]
dist =
from P3 to P4
dist = sqrt [ (x4-x3)^2 +(y4-y3)^2 ]
dist =
dist = sqrt [ (x1-x4)^2 +(y1-y4)^2 ]
dist =
Perimeter is exactly = +
+
+
To find the area we have to use the non-right triangle formula of
1/2*a*b*sin(x)
if we knew the angle at P1 then we could find the area of the triangle that is made up by P1, P2 and P4
P1(1,1) (x1,y1)
P2(6,2) (x2,y2)
P3(5,4) (x3,y3)
P4(2,5) (x4,y4)
let's use that fancy slope formula to find the angle
tan(Ф) = abs [ (m1-m2)/(1+m1.m2) ]
m1 = P1P2
m1 = y2-y1 / x2-x1
m1 = 1/5 or .2
m2 = P1P4
m2 = y4-y1 / x4-x1
m2 = 4
now plug into the formula for the angle of P1
tan(Ф) = abs [(0.2 - 5) / (1+0.2*5)]
Ф = arcTan(2.4)
Ф=-67.380135°
the negative signs just means going from P1P4 to P1P2 is -67.380135° we can take it to be positive as well
now plug that angle into our area formula for a non-right triangle
1/2*a*b*sin(x)
area of triangle P1,P2,P4 = 1/2**
*sin(67.380135°)
area P1,P2,P4 = 9.703290477
now lets find that other triangle up at the top formed by P2, P3, P4
let's find the slope of the two legs we know the lengths of, of the part that looks a little like a "hat" on top of our 1st triangle
P1(1,1) (x1,y1)
P2(6,2) (x2,y2)
P3(5,4) (x3,y3)
P4(2,5) (x4,y4)
m3= P4P3
m3=y3-y4 / x3-x4
m3 = -1/3
m4 = P3P2
m4 = y2-y3 / x2-x3
m4 = -2/1
m4 = -2
now find the angle between them
tan(Ф) = abs [ (m3-m4)/(1+m3.m4) ]
Ф = arcTan(1)
Ф = 45°
this is the other angle on the 180 line so we have 180-45 =135 as the angle inside our triangle at P3, although, sin 135 and 45 are going to be the same.
now plug that info into our special area of a non-right triangle
area of P2P3P4 = 1/2**
*sin(135)
area of P2P3P4 = 2.5
2.5 + 9.703290477 = 12.20329048
area =12.20329048
The feasible region is shown in the picture above. The horizontal axis is the X axis.
If you now draw a family of lines parallel to through the region, increasing $C$, that is, the X-intercept, we will have the maximum value of $C$ when the line passes through the endpoint $(-2,6)$ .
Hence, $C=5(-2)+4(6)=14$
