Corey drives 50 miles & the road is 200 miles, & he wastes 30 miles per gallon, how many more gallons to get 200 miles ?

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:

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

(In this kind of problems always the additional equation is the restricion, in this case, V=500).

Let's divide the first and second equations by π:

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

Isolate λ from the second equation:

$\lambda =\frac{0.06}{r}$

Isolate h from the third equation:

$h=\frac{500}{\pi r^2}$

And then, replace λ and h in the first equation:

$0.06*\frac{500}{\pi r^2} +0.2r=2*(\frac{0.06}{r})r\frac{500}{\pi r^2} \\\frac{30}{\pi r^2}+0.2r= \frac{60}{\pi r^2}$

Multiply all the resultant equation by $\pi r^{2}$ :

$30+0.2\pi r^3=60\\0.2\pi r^3=30\\r^3=\frac{30}{0.2\pi } =\frac{150}{\pi}\\r=\sqrt[3]{\frac{150}{\pi}}\approx 3.628cm$

Then, find h by the equation $h=\frac{500}{\pi r^2}$ founded above:

$h=\frac{500}{\pi r^2}\\h=\frac{500}{\pi (3.628)^2}=12.093cm$

4 0

3 years ago

Given one of the roots of the quadratic equation x2 + kx + 12 = 0 is three times the other root. Find the values of k.

gtnhenbr [62]

$\huge\bold{\purple{\bold{⚡Kaboom!⚡}}}$

5 0

2 years ago

4,792÷8 show your work

Svetradugi [14.3K]

Answer:

Step-by-step explanation:

Look at the picture.

Use the long division.

3 0

3 years ago

Helllp im doing my exam !

Tcecarenko [31]

log x ( 5 × 12 ) = 7

x^7 = 60

..................................

7 0

3 years ago

Whats wrong in this equation ?

Inga [223]

The second step is wrong. What should've been done is to find greatest common factor (gcf) of 1/6 and -2. This is because you cannont add together a number with a variable to a number without a variable. So get the variable by itself by subtracting 1/6 from both sides.

1/5x + 1/6 = -2
___— 1/6_— 1/6
____________

Turn -2 into a fraction and find the gcf of -2 and 6:

1/5x + 1/6 = -2
___— 1/6_— 1/6
____________

1/5x = -2/1 — 1/6 ——> 1/5x = -12/6 — 1/6
1/5x = -13/6

Then divide each side by 1/5 to get the variable by itself; remeber: when dividing a fraction by a fraction, you multiply by the reciprocal.

5/1 • 1/5x = -13/6 • 5/1
x = 65/6

Then, simplify

65/6} 10.83 or 10 83/100

6 0

3 years ago