?'s questions at Yahoo Answers regarding optimization

[MarkFL]

Here are the questions:

What is the function and answer to these optimization questions?

1. A piece of wire 40 cm long is to be cut into two pieces. One will be bent to form a circle; the other will be bent to form a square. Find the lengths of the two pieces that cause the sum of the area of the circle and the area of a square to be a minimum.

2. A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum value?

3. The concentration in mg/cm^3, of a particular drug in a patient's bloodstream is given by the formula C(t) = (0.12t) / (t^2+2t+2), where t represents the number of hours after the drug is taken.
a) Find the maximum concentration on the interval 0<=t<=4.
b) Determine when the maximum concentration occurs

I have posted a link to this topic so the OP can see my work.

 

[MarkFL]

Re: ?'s questions at Yahoo Answers regarding optimization

Hello ?,

1.) A piece of wire 40 cm long is to be cut into two pieces. One will be bent to form a circle; the other will be bent to form a square. Find the lengths of the two pieces that cause the sum of the area of the circle and the area of a square to be a minimum.

Let's let the length of the wire be $0<L$, and the let $0\le S$ be the portion of the wire used to form the square, and let the portion of the wire that is the circumference of the circle be $0\le C$

Each side of the square will be $\dfrac{S}{4}$, and so the area of the square is:

\(\displaystyle A_S=\left(\frac{S}{4} \right)^2=\frac{S^2}{16}\)

The area of the circle is:

\(\displaystyle A_C=\pi r^2\)

Now, the circumference of the circle is $C$ and so the radius of the circle is:

\(\displaystyle r=\frac{C}{2\pi}\)

hence the area of the circle in terms of $C$ is:

\(\displaystyle A_C=\pi\left(\frac{C}{2\pi} \right)^2=\frac{C^2}{4\pi}\)

Adding the two areas, we find the total area of the two shapes is:

\(\displaystyle A=A_S+A_C=\frac{S^2}{16}+\frac{C^2}{4\pi}\)

To find the absolute extrema, we first observe we must have:

$0\le S\le L$ which also means $0\le C\le L$

So, we will find the local extrema within this interval, and also look at the area at the end-points of the interval as possible absolute extrema as well.

There are 3 ways to find the local extremum, since the area is a quadratic in $S$.

For the first two methods, we wish to have an area function in one variable, and so we may use:

$C=L-S$

and we have:

\(\displaystyle A(S)=\frac{S^2}{16}+\frac{(L-S)^2}{4\pi}\)

Method 1: find the axis of symmetry

We will write the area function in standard quadratic form:

\(\displaystyle A(S)=\frac{4+\pi}{16\pi}S^2-\frac{L}{2\pi}S+\frac{L^2}{4\pi}\)

Thus, the axis of symmetry is at:

\(\displaystyle S=-\frac{-\frac{L}{2\pi}}{2\left(\frac{4+\pi}{16\pi} \right)}=\frac{4L}{4+\pi}\)

Since the parabola opens upward, we know the global minimum is at this value of $S$.

Thus, this is the amount of the wire that should go to the square to minimize the total area. Analysis of the end-points reveals:

\(\displaystyle A(0)=\frac{L^2}{4\pi}\)

\(\displaystyle A(L)=\frac{L^2}{16}\)

Since $A(0)>A(L)$, we find that to maximize the area, none of the wire should go to the square, and this makes sense as a circle will enclose more area per perimeter than a square.

Method 2: Equate derivative of area function to zero and solve for $S$ to get critical value.

\(\displaystyle A(S)=\frac{S^2}{16}+\frac{(L-S)^2}{4\pi}\)

\(\displaystyle A'(S)=\frac{S}{8}-\frac{L-S}{2\pi}=\frac{\pi S-4(L-S)}{8\pi}=0\)

This implies:

\(\displaystyle \pi S-4(L-S)=0\)

\(\displaystyle S=\frac{4L}{4+\pi}\)

To determine the nature of the associated extremum, we may use either the first or second derivative tests.

First derivative test:

\(\displaystyle A'(0)=\frac{\pi(0)-4(L-0)}{8\pi}=-\frac{L}{2\pi}<0\)

\(\displaystyle A'(L)=\frac{\pi L-4(L-L)}{8\pi}=\frac{L}{8}>0\)

Thus, the first derivative test shows the extremum is a minimum.

Second derivative test:

\(\displaystyle A''(S)=\frac{d}{dS}\left(\frac{\pi S-4(L-S)}{8\pi} \right)=\frac{p+4}{8\pi}>0\)

This shows the area function is concave up everywhere, and thus the extremum must be a global minimum.

Method 3: Use optimization with constraint

We have the objective function:

\(\displaystyle A(C,S)=\frac{S^2}{16}+\frac{C^2}{4\pi}\)

subject to the constraint:

\(\displaystyle g(C,S)=C+S-L=0\)

Using Lagrange multipliers, we find:

\(\displaystyle \frac{C}{2\pi}=\lambda\)

\(\displaystyle \frac{S}{8}=\lambda\)

this implies:

\(\displaystyle \frac{C}{2\pi}=\frac{S}{8}\)

\(\displaystyle C=\frac{\pi S}{4}\)

Substituting this into the constraint, we find:

\(\displaystyle \frac{\pi S}{4}+S-L=0\)

Solving for $S$, we find:

\(\displaystyle S=\frac{4L}{4+\pi}\)

The remaining analysis is the same as for Method 1.

In conclusion, we have found to minimize the area, we need:

\(\displaystyle S=\frac{4L}{4+\pi}\)

\(\displaystyle C=\frac{\pi}{4}S=\frac{\pi}{4}\cdot\frac{4L}{4+\pi}=\frac{\pi L}{4+\pi}\)

Using the given value $L=40\text{ cm}$ we find then:

\(\displaystyle S=\frac{160}{4+\pi}\,\text{cm}\)

\(\displaystyle C=\frac{40\pi}{4+\pi}\,\text{cm}\)

 

[MarkFL]

Re: ?'s questions at Yahoo Answers regarding optimization

2.) A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum value?

I am assuming two things here...we wish to maximize the volume and the sirface area is the outer surface area only.

Let's let $0\le L$ be the lengths of the sides of the square base and $0<S$ be the outer surface area. We will let $0\le h$ be the height of the box. We may then write:

(1) \(\displaystyle V(h,L)=hL^2\)

(2) \(\displaystyle S(h,L)=L^2+4hL\)

We should observe that we require:

\(\displaystyle 0\le L\le\sqrt{S}\)

Method 1: Single variable calculus

Solving (2) for $h$, we find:

\(\displaystyle h=\frac{S-L^2}{4L}\)

Substituting for $h$ into (1) we have:

\(\displaystyle V(L)=\left(\frac{S-L^2}{4L} \right)L^2=\frac{1}{4}\left(SL-L^3 \right)\)

Differentiating with respect to $L$ and equating to zero, we find:

\(\displaystyle V'(L)=\frac{1}{4}\left(S-3L^2 \right)=0\)

And this implies:

\(\displaystyle S-3L^2=0\)

Taking the positive root since $L$ represents a measurement of length, we have:

\(\displaystyle L=\sqrt{\frac{S}{3}}\)

and so:

\(\displaystyle h=\frac{S-\left(\sqrt{\frac{S}{3}} \right)^2}{4\sqrt{\frac{S}{3}}}= \frac{1}{2}\sqrt{\frac{S}{3}}\)

To determine the nature of the extremum associated with this critical value,we may use the first and second derivative tests.

First derivative test:

\(\displaystyle V'(0)=\frac{1}{4}\left(S-3(0)^2 \right)=\frac{S}{4}>0\)

\(\displaystyle V'(\sqrt{S})=\frac{1}{4}\left(S-3(\sqrt{S})^2 \right)=-\frac{S}{2}<0\)

Thus, the first derivative test shows the extremum is a maximum. Since we have:

\(\displaystyle V(0)=V(\sqrt{S})=0\), we know this is the absolute maximum on the restricted domain of $L$.

Second derivative test:

\(\displaystyle V''(L)=\frac{d}{dL}\left(\frac{1}{4}\left(S-3L^2 \right) \right)=-\frac{6L}{4}\)

Thus, on the restricted domain, the volume function is concave down, hence the extremum is a maximum.

Method 2: Multi-variable calculus (optimization with constraint)

We have the objective function:

\(\displaystyle V(h,L)=hL^2\)

subject to the constraint:

\(\displaystyle g(h,L)=L^2+4hL-S=0\)

Using Lagrange multipliers, we find:

\(\displaystyle L^2=4L\lambda\)

\(\displaystyle 2hL=(2L+4h)\lambda\)

This implies:

\(\displaystyle \lambda=\frac{L^2}{4L}=\frac{hL}{L+2h}\)

\(\displaystyle \frac{L^2(L+2h)-4hL^2}{4L(L+2h)}=0\)

\(\displaystyle \frac{L^2(L-2h)}{4L(L+2h)}=0\)

Thus, knowing $L=0$ is a minimum, we are left with:

\(\displaystyle L=2h\)

and substitution of this into the constraint reveals:

\(\displaystyle (2h)^2+4h(2h)-S=0\)

\(\displaystyle h=\frac{1}{2}\sqrt{\frac{S}{3}}\)

and so:

\(\displaystyle L=2h=\sqrt{\frac{S}{3}}\)

So, in conclusion, we have found that the volume of the box is maximized for:

\(\displaystyle L=\sqrt{\frac{S}{3}}\)

\(\displaystyle h=\frac{1}{2}\sqrt{\frac{S}{3}}\)

Using the given value of \(\displaystyle S=108\text{ in}^2\), we then have:

\(\displaystyle L=6\text{ in}\)

\(\displaystyle h=3\text{ in}\)

 

[MarkFL]

Re: ?'s questions at Yahoo Answers regarding optimization

3.) The concentration in \(\displaystyle \frac{\text{mg}}{\text{cm}^3}\), of a particular drug in a patient's bloodstream is given by the formula \(\displaystyle C(t)=\frac{0.12t}{t^2+2t+2}\), where $t$ represents the number of hours after the drug is taken.

a) Find the maximum concentration on the interval $0\le t\le4$.

b) Determine when the maximum concentration occurs.

First, we want to find the critical values by equating the derivative of the concentration function to zero. Use of the quotient rule gives:

\(\displaystyle C'(t)=\frac{0.12(t^2+2t+2)-0.12t(2t+2}{(t^2+2t+2)^2}=\frac{0.12(2-t^2)}{(t^2+2t+2)^2}=0\)

The denominator has no real roots, and for the given interval the only critical value is then:

\(\displaystyle t=\sqrt{2}\)

To determine the nature of the extremum associated with this critical value, we may use the first and second derivative tests.

First derivative test:

\(\displaystyle C'(0)=\frac{0.12(2-0^2)}{(0^2+2(0)+2)^2}=0.06>0\)

\(\displaystyle C'(4)=\frac{0.12(2-4^2)}{(4^2+2(4)+2)^2}=-\frac{21}{8450}<0\)

Thus, the first derivative test shows that the extremum is the absolute maximum.

Second derivative test:

\(\displaystyle C''(t)=\frac{d}{dt}\left(\frac{0.12(2-t^2)}{(t^2+2t+2)^2} \right)=0.12\left(\frac{(t^2+2t+2)^2(-2t)-(2-t^2)(2(t^2+2t+2)(2t+2))}{(t^2+2t+2)^4} \right)=\frac{0.24(t^3-6t-4)}{(t^2+2t+2)^3}\)

Knowing the denominator is positive for all real $t$, we need only look at:

\(\displaystyle (\sqrt{2})^3-6(\sqrt{2})-4=-4\sqrt{2}-4<0\)

And so we know the extremum is a maximum. Thus, we have found:

a) The maximum concentration in \(\displaystyle \frac{\text{mg}}{\text{cm}^3}\) is

\(\displaystyle C_{\max}=C(\sqrt{2})=\frac{3}{50}(\sqrt{2}-1)\approx0.0248528137423857\)

b) This maximum occurs for $t$:

\(\displaystyle t=\sqrt{2}\text{ hr}\approx1\text{ hr }24\text{ min }51.1688245432\text{ sec }\).

 

[CellCoree]

1. The function in this optimization question is the sum of the areas of the circle and square, which can be represented as A = πr^2 + s^2, where r is the radius of the circle and s is the length of one side of the square. The constraint is that the total length of the wire is 40 cm, so we can write this as 2πr + 4s = 40. We can solve for one variable in terms of the other, such as r = (40-4s)/(2π), and substitute into the first equation to get A = π[(40-4s)/(2π)]^2 + s^2 = 800/π + s^2 - 16s/π. We can then take the derivative with respect to s and set it equal to 0 to find the critical point. After solving for s, we can plug it back into the original equation to find the corresponding value of r. This will give us the lengths of the two pieces of wire that will produce the minimum sum of areas.

2. The function in this optimization question is the surface area of the box, which can be represented as A = s^2 + 4s^2 = 5s^2. The constraint is that the surface area is 108 square inches, so we can write this as s^2 + 4s^2 = 108. We can solve for one variable in terms of the other, such as s = √(108/5), and substitute into the first equation to get A = 108/5 + 108/5 = 216/5. This gives us the maximum surface area for the given dimensions of the box.

3. a) The maximum concentration in this optimization question can be found by taking the derivative of the concentration formula with respect to t and setting it equal to 0. After solving for t, we can plug it back into the original equation to find the maximum concentration on the given interval.

b) To determine when the maximum concentration occurs, we can take the second derivative of the concentration formula and evaluate it at the critical point found in part a. If the second derivative is negative, then the critical point is a maximum; if it is positive, then the critical point is a minimum. This will tell us when the maximum concentration occurs within the given time interval.