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# Calculus I: Lesson 10: Max and Min Problems 3

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Calculus I

Lesson 10: Max and Min Problems 3

Example 1: Paper Folding

Problem: A sheet of paper 4 inches wide by 8 inches high is folded so that the bottom right corner of the sheet touches the left hand edge of the sheet. The tip of the corner is no more than 4 inches above the bottom edge of the paper. Then the paper is creased (see figure). Find the length L of the crease, and find how to fold the paper so that L is minimum.

Solution:

Let h, x, and y be as shown in the diagram below.

> restart;

> A1:=plots[textplot]({[2.6,1.9,'L'],[3.7,2,'h'],[.1,1.1,`y`],[.5,.3,`x`]},align=RIGHT):

> A3:=plot({[[0,0],[0,8],[4,8],[4,4.45],[1.219,0],[0,0]],[[4,4.45],[0, 2.5],[1.219,0]]},color=blue):

> A4:=plot([[4,4.45],[4,0],[1.219,0]],color=red):

> A5:=plots[polygonplot]([[4,4.45],[0,4.45],[0,2.5]],style=patch,color=tan):

> plots[display]([A1,A3,A4,A5],axes=boxed,scaling=constrained); We can see several equations relating x, y, h, and L in the diagram. For example, the small right triangle with legs x and y has a hypotenuse which is 4-x units long. This gives eq1.

> eq1 := y^2+x^2=(4-x)^2; eq2 comes from the right triangle with hypotenuse L and legs h and 4-x.

> eq2 := L^2 = (4-x)^2 + h^2; Now it is easy to work out that the tan right triangle with hypotenuse h and legs 4 and h-y is similar to the right triangle with hypotenuse 4-x and corresponding legs y and x. So we get eq3.

> eq3 := 4/(h-y)=y/x; > h := solve(eq3,h); > x := solve(eq1,x); > L := unapply(sqrt(op(2,simplify(eq2))),y); So we have L expressed as a function of one variable y. Examining the behavior of L as y varies,

> plot(L,2..4); we see that there is a minimum length crease of about inches at about inches. We can get a more precise value with fsolve, by locating the x between 2.6 and 3 where the derivative of the crease function is 0.

> y1 := fsolve(diff(L(y),y),y,2.6..3); Now check the value of x and L for this value of y.

> minL := L(y1); > minx := subs(y=y1,x); Notice the nice integer value of minx . This gives rise to a simple construction of the crease of minimum length: by folding twice along the bottom edge, we can mark the point 1 inch from the left edge of the paper. Then bring the corner point up to the left edge and crease.

Exercise: Use Maple to draw the diagram showing the minimum length crease.

Exercise: Suppose we wanted to minimize L + y, rather than just L. Would the minimum occur at the same place? Work it out.

Example 2

A woman's blood pressure p ranges from a high (the systolic pressure) to a low

(the diastolic pressure) so that at time t in seconds, p is given (approximately) by

p = P(t) = 100 + 20 sin(5t).

The time between two successive high pressures is the time needed for one heartbeat.

First, plot P(t).

a) Find an expression for the rate at which the woman's blood pressure is changing

at time t.

b) At what time is the woman's blood pressure a max? a min?

c) What is the woman's max blood pressure? min blood pressure?

d) How many times per minute does the woman's heart beat?

> restart: with(plottools): with(plots):

```Warning, the name changecoords has been redefined
```

> f:= 100 + 20*sin(5*x); > plot(f(x), x = 0..10); > f:= x -> 100 + 20*sin(5*x); > D(f); a) The womans blood pressure changes at a rate of .

> fsolve(100 * cos( 5*x) = 0, x, 0..1); > fsolve(100* cos(5*x) = 0, x, .5..1); > fsolve(100* cos(5*x) = 0, x, 1..2); b) Since f takes on max min when , we see that

f has a max when x is about .3141592654 and 1.570796327.

Also, f has a min when x is about .942477796.

We otained these estimates by setting .

> f(.3141592654); > f(.9424777961); > f(1.570796327); c) We see the max blood pressure is 120 and the minimum is 80.

> 1.570796327 - .3141592654; > 60 / 1.256637062; d) Each heartbeat takes approx 1.256637062 seconds.

She has approximately 48 heartbeats per minute.

Example 3

A triangle is formed in the first quadrant by a tangent line to the graph of the curve and the coordinate axes. What is the area of the largest such triangle? First let's see a diagram of the situation.

> f := x->exp(-x);
a1 := 1.5: a2 := 2.5:
eqTanLine := (x,a) -> f(a) + (x-a) * D(f)(a):
b1 := solve( eqTanLine(x,a1)=0, x):
b2 := solve( eqTanLine(x,a2)=0, x):
graph := plot( f(x), x = 0..5, thickness=2):
tanLine1 := plot( eqTanLine(x,a1), x = 0..b1, color=blue, thickness=2):
tanLine2 := plot( eqTanLine(x,a2), x = 0..b2, color=green, thickness=2):
line1 := line( [0, 0],[b1, 0], color=blue, thickness=2):
line2 := line( [0, 0],[0, eqTanLine(0,a1)], color=blue, thickness=2):
line3 := line( [0, 0],[b2,0], color=green, thickness=2):
line4 := line( [0, 0],[0, eqTanLine(0,a2)], color=green, thickness=2):
display( graph, tanLine1, tanLine2, line1, line2,line3, line4);  From the above plot we see that the triangle changes with our choice of tangent lines.

We need to find a tangent line that gives a triangle of maximal area.

Find the x-intercept of a tangent line.

> solve( exp(-a)-exp(-a)*(x-a)=0,x); So, the x-intercept of the tangent line when x = a is: 1 + a.

Find the y-intercept of a tangent line.

> tl(0,a); So, the y-intercept of the tangent line when x = a is: Thus using the tangent line for x = a we have that the triangle has

height = and

base = 1 + a. We can now calculate the area of the triangle.

> A:= 0.5 * (exp(-a)+exp(-a)*a) * ( 1 + a); > plot(A(a), a = 0..5); From the plot of the area function (area of the triangle), we see that there is indeed

a maximum. We need to precisely locate for what a values does this maximum occur.

At this maximum . So, we solve .

> AA:= a -> 0.5 * (exp(-a)+exp(-a)*a) * ( 1 + a); > diff(AA(a),a); > solve(% = 0, a); > plot(A(a), a = -2..3); From the plot, we see that indeed for a = -1,1.

We are only interested in positive values of a.

Thus the answer is that the area of the triangle is largest when

the tangent line is for x = 1.

Example 4

A cone is to be formed by revolving a right triangle with a hypotenuse of length 3

feet about one of its legs. What is the volume of the largest cone that can be formed

in this manner? See the diagram below.

> c := cone([0,0,4], sqrt(8), -4, color=green):
t := pointplot3d([[0,0,0], [sqrt(8),0,0], [0,0,4],[0,0,0]], connect=true, color=black, thickness=2):
ground := plot3d(0,x=-3..3,y=-3..3,color=grey,grid=[5,10], style=patchnogrid):
display([c,t,ground],scaling=constrained,style=wireframe, orientation=[40,60]); Let a denote the length of the base of the right triange, and b denote the

height of the right triangle. We are given the hypotenuse is 3 feet.

The volume of the resulting cone is: . We want to maximize the volume.

Since we have a right triangle we know that .

> restart:

> evalf(Pi); > V:= (1/3) * (3.141592654) * (9 - b^2) * b; > plot(V(b), b = 0..3); From the estiamted plot, we see that the volume has one maximum and that

at this maximum . Hence, we solve: .

> VV:= b -> (1/3) * Pi * (9 - b^2) * b; > diff(VV(b),b); > solve(%=0,b); > evalf(sqrt(3)); The cone will have maximum volume when b = sqrt(3) and = 9 - 3 = 6, i.e, and .