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.
Answer: 33.75
Step-by-step explanation:
It’s right
Answer:
for the first one I think it is -5/-2 second is 1/-4 third one is 2/1 this may not be correct
Step-by-step explanation:
Answer:
Step-by-step explanation:
We are given that a function
We have to find the average value of function on the given interval [0,7]
Average value of function on interval [a,b] is given by
Using the formula
![f_{avg}=\frac{1}{7}[e^{\frac{x}{7}}\times 7)]^{7}_{0}](https://tex.z-dn.net/?f=f_%7Bavg%7D%3D%5Cfrac%7B1%7D%7B7%7D%5Be%5E%7B%5Cfrac%7Bx%7D%7B7%7D%7D%5Ctimes%207%29%5D%5E%7B7%7D_%7B0%7D)
By using the formula
Because 
Hence, the average value of function on interval [0,7]
The bottom one is correct.
