<span>Constraints (in slope-intercept form)
x≥0,
y≥0,
y≤1/3x+3,
y</span>≤ 5 - x
The vertices are the points of intersection between the constraints, or the outer bounds of the area that agrees with the constraints.
We know that x≥0 and y≥0, so there is one vertex at (0,0)
We find the other vertex on the y-axis, plug in 0 for x in the function:
y <span>≤ 1/3x+3
y </span><span>≤1/3(0)+3
y = 3.
There is another vertex at (0,3)
Find where the 2 inequalities intersect by setting them equal to each other
(1/3x+3) = 5-x Simplify Simplify Simplify
x = 3/2
Plugging in 3/2 into y = 5-x: 10/2 - 3/2 = 7/2
y=7/2
There is another vertex at (3/2, 7/2)
There is a final vertex where the line y=5-x crosses the x axis:
0 = 5 -x , x = 5
The final vertex is at point (5, 0)
Therefore, the vertices are:
(0,0), (0,3), (3/2, 7/2), (5, 0)
We want to maximize C = 6x - 4y.
Of all the vertices, we want the one with the largest x and smallest y. We might have to plug in a few to see which gives the greatest C value, but in this case, it's not necessary.
The point (5,0) has the largest x value of all vertices and lowest y value.
Maximum of the function:
C = 6(5) - 4(0)
C = 30</span>
To solve this, first we'll find the area of the rectangle A,
Area=length × width
?=24m×20m
480m=24m×20m
480m squared=area of the rectangle A
now we'll find the width of rectangle B,
"the width of rectangle B is 12 meters less than the width of rectangle A",
20m-12m= 8m
8m=width of rectangle B
finally we'll find the length of rectangle B,
area of the rectangle B= 480msquared
width= 8m
length=? (to find this divide the area by the width)
480÷8=60m
length of the rectangle B=60m
Answer: 12.6 m
Step-by-step explanation:
Use the pythagorean theorem, we get that the height, or x, is
sqrt(22x22-18x18) = sqrt(484-324) = sqrt(160) = 4sqrt(10) = 12.6 m
