A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the cost to make the can if the metal for the sides will cost $1.25 per 2 cm and the metal for the bottom will cost $2.00 per 2 cm ?

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V=

According to the problem,

The surface area of the base of the can is =

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× )

The lateral surface area of the can is =

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× )

Total cost of metal is C= 2.00 +

Putting

Differentiating with respect to r

Again differentiating with respect to r

To find the minimize cost, we set C'=0

⇒r=1.71

Now,

When r=1.71 cm, the metal cost will be minimum.

Therefore,

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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