Using the graph, what is the value of x that will produce the maximum volume? f(x)=4x^3-20x^2+24x?

Answer 1

doing this yourself will actually help you to learn the mathematics, instead of asking someone else to do it for you = will never learn.

Answer 2

permit x represent the two comparable aspects of the rectangle, or the width.. permit y represent the single area, or length. the fringe would be 2x+y = 40 4. Now locate y in terms of x. y = 40 4 - 2x. you have mentioned the fee of x that produces the optimal, i'll assume you propose section. section is represented via length situations width. The width is represented via x, the dimensions via y, yet y = 40 4 - 2x. So section = x(40 4-2x) section = 44x - 2x^2 Now use differentiation to locate the fee that x would be for optimal section. d/dx(44x-2x^2) = 40 4-4x. x would be a optimal while the gradient equals 0, so permit the by-product equivalent 0. 40 4-4x = 0 40 4 = 4x x = 11 the portion of the backyard would be optimal while x, or width equals 11ft. to locate y, 40 4 - 2x = y, so y equals 22 ft.

Answer 3

You can do this by USING THE GRAPH, like the problem states, or you can use calculus. If you were to use calculus, you first would take the derivitive which would give you 12x^2 - 40x + 24. Setting that equal to 0 and solving for X using the quadratic formula would give you the value of X that you're looking for.

Answer 4

The exact volume is 8.45

Let me explain it to you clearly.
f(x) = 4x^3-20x^2+24x

dy/dx = 12x^2-40x+24
when dy/dx is o find the f(x) value

4(3x^2-10x+6) = 0

3x^2-10x+6 = 0

use -b+or-Square root(b^2-4ac)/2a

dy/dx becomes 0 at 2 places

1st when x =0.784745 and the next at x =2.5485837

volume is maximum when d^2y/dx^2 is negative

so d^2y/dx^2=24x-40
when x is 0.784745
d^2y/dx^2 = -21.16

when x=2.5485837
thrn d^2y/dx^2 = 21.152


Therefore the volume is maximum at x = 0.784745 since d^2y/dx^2(second derivative) is negative which gives out a maximum.

so the maximum volume is obtained by substituting x=0.784745 to
f(x) =8.45

maximum volume is 8.45

Answer 5

f(x)=4x^3-20x^2+24x

First derivative of f(x) with respect to x =f1(x)= 12x^2 - 40x + 24

for any function to have maximum or minimum one need to have first derivative = 0

so solving f1(x) =0
12x^2 - 40x + 24 = 0
=> 4(x^2 - 10x + 6) = 0
=> x = (10 +sqrt(10^2 - 4*1*6))/2 ,
and (10 +sqrt(10^2 - 4*1*6))/2

i.e at x = (5+sqrt19)/2 , (5-sqrt19)/2 f1(x) = 0

putting them in f(x) we will get maximum or minimum values
at x = (5+sqrt19)/2 f(x)= -2.52
at x = (5-sqrt19)/2 f(x)= 9.45

so we have maximum value at (5-sqrt19)/2 equal to 9.45.

Answer 6

To find maximums or minimums, you need to find when the derivative, f'(x)=0

f(x)=4x^3-20x^2+24x
f'(x)=12x^2-40x+24
0=12x^2-40x+24
x=.78 and 2.55 one of these is the minimum and the other is the maximum, put them back into the original equation to see which is the maximum

f(.785)= 4(.78)^3- 20(.78)^2+ 24(.78)= 9.45
f(2.548)= 4(2.55)^3- 20(2.55)^2+ 24(2.55)= -2.52

There is a minimum at 2.55,-2.52
The maximum is at .78,9.45