F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)
d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.
This is a very long question. I'm not going to write all of it out but I will give you a starting point. Find your x by making y in the formula equal to 0.
2x + 3y = 1470
2x + 3(0) = 1470
2x = 1470
x = 735
Your furthest point on the x axis is (735,0).
Do the same for y.
2x + 3y = 1470.
2(0) + 3y = 1470
3y= 1470
y= 490
Your highest point is (0,490).
Now that both are plotted, draw a straight line connecting the two points. There's your graph.
Check
Answer:
r=3
Step-by-step explanation:
3(r - 7) = 4(2 - 2r) + 4
Distribute
3r -21 = 8 -8r +4
Combine like terms
3r -21 = 12 -8r
Add 8r to each side
3r +8r-21 = 12 -8r+8r
11r -21 = 12
Add 21 to each side
11r -21+21 =12+21
11r = 33
Divide each side by 11
11r/11 = 33/11
r = 3
