If f is the function given by f(x)= integral from (4 to 2x) of sqrt(t^2-t)dt then f'(2) =?

2 answers:

5 0

The question ask to calculate the function f'(2) in the F is the function given by f(x) = integral from (4 to 2x) of sqrt(t^2-t), base on that, the possible answer would be that the f(2) is 4 sqrt 3. I hope you are satisfied with my answer and feel free to ask for more if you have questions

sammy [17]2 years ago

3 0

Hello,

$f(x)= \int\limits^{2x}_4 {\sqrt{t^2-t} \, dt \\

 \dfrac{df(x)}{dx} = \dfrac{df(x)}{d(2x)} * \dfrac{d(2x)}{dx} =2* \sqrt{(2x)^2-2x} \\

f'(2)=2 * \sqrt{4^2-4} =2* \sqrt{12} =4 \sqrt{3}$

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A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. The material for the sides of the can costs

iogann1982 [59]

Answer:

r=3.628cm

h=12.093cm

Step-by-step explanation:

For this problem we are going to use principles, concepts and calculations from multivariable calculus; mainly we are going to use the Lagrange multipliers method. This method is thought to help us to find a extreme value of a multivariable function 'F' given a restriction 'G'. F represents the function that we want to optimize and G is just a relation between the variables of which F depends. The Lagrange method for just one restriction is:

$\nabla F=\lambda \nabla G$

First, let's build the function that we want to optimize, that is the cost. The cost is a function that must sum the cost of the sides material and the cost of the top and bottom material. The cost of the sides material is the unitary cost (0.03) multiplied by the sides area, which is $A_s=2\pi rh$ for a cylinder; while the cost of the top and bottom material is the unitary cost (0.05) multiplied by the area of this faces, which is $A_{TyB}=2\pi r^2$ for a cylinder.

So, the cost function 'C' is:

$C=2\pi rh*0.03+2\pi r^2*0.05\\C=0.06\pi rh+0.1\pi r^2$

The restriction is the volume, which has to be of 500 cubic centimeters:

$V=500=\pi r^2h\\500=\pi hr^2$

So, let's apply the Lagrange multiplier method:

$\nabla C=\lambda \nabla V\\\frac{\partial C}{\partial r}=0.06\pi h+0.2\pi r\\\frac{\partial C}{\partial h}=0.06\pi r\\\frac{\partial V}{\partial r}=2\pi rh\\\frac{\partial V}{\partial h}=\pi r^2\\(0.06\pi h+0.2\pi r,0.06\pi r)=\lambda (2\pi rh,\pi r^2)$

At this point we have a three variable (h,r, λ)-three equation system, which solution will be the optimum point for the cost (the minimum). Let's write the system:

$0.06\pi h+0.2\pi r=2\lambda \pi rh\\0.06\pi r=\lambda \pi r^2\\500=\pi hr^2$