Question

1 optimization calculus question, 50 pts please help

Answer 1

This is about optimization problems in mathematics.

Dimensions; Height = 48 inches; Radius = 48/π inches

We are told the combined length and girth is 144 inches.

Girth is same as perimeter which is circumference of the circular side.

Thus; Girth = 2πr

If length of cylinder is h, then we have;

2πr + h = 144

h = 144 - 2πr

Now, to find the dimensions at which the max volume can be sent;

Volume of cylinder; V = πr²h

Let us put 144 - 2πr for h to get;

V = πr²(144 - 2πr)

V = 144πr² - 2π²r³

Differentiating with respect to r gives;

dV/dr = 288πr - 6π²r²

Radius for max volume will be when dV/dr = 0

Thus; 288πr - 6π²r² = 0

Add 6π²r² to both sides to get;

288πr = 6π²r²

Rearranging gives;

288/6 = (π²r²)/πr

48 = πr

r = 48/π inches

Put 48/π for r in h = 144 - 2πr to get;

h = 144 - 2π(48/π)

h = 144 - 96

h = 48 inches

Explanation:

Answer 2

radius: 14.96 in

length: 47 in

Explanation:

The dimensions of the package with maximum volume can be found by differentiating the volume function, subject to the constraint on the dimensions.

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volume function

The volume of the cylindrical package is ...

V = πr²h

The constraint on the dimensions is ...

circumference + length = 141 inches

2πr +h = 141 . . . . . at maximum volume

Solving the second equation for h, we can write the volume function in terms of r alone:

h = 141 -2πr

V = πr²(141 -2πr) . . . . substitute for h

V = 141πr² -2π²r³ . . . eliminate parentheses

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derivative

Differentiating with respect to radius, we find the radius at maximum volume must satisfy ...

V' = 282πr -6π²r² = 0

Dividing by 6πr, we can simplify this to ...

47 -πr = 0

r = 47/π ≈ 14.96 . . . . inches (radius)

h = 141 -2πr = 47 . . . inches (length)