Optimization Problems in Physics

Solved Problems

Solution.

We use Newton's Second Law to write the motion equation for the body in vector form:

where \(\vec{a}\) is the acceleration of the box, \(m\) is its mass, \(\vec{F}_f\) is the friction force, \(m\vec{g}\) is the weight of the body, and \(\vec{N}\) is the normal force.

The coordinate system is chosen so that the \(x\)-axis is parallel to the inclined plane and the \(y\)-axis is orthogonal to the \(x\)-axis.

In components form, the above equation becomes

As

we get

Hence, the magnitude (the absolute value) of the acceleration is given by

The distance and the velocity are defined by the equations

where \(a \gt 0.\)

The time of movement is

Then the distance travelled for this time is equal to

Substituting the acceleration \(a,\) we obtain the function \(s\left( \alpha \right):\)

The derivative of \(s\left( \alpha \right)\) is given by

Determine the critical value of \(\alpha:\)

Using the First Derivative Test, we can show that the angle \(\alpha = \arctan \frac{1}{k}\) corresponds to the smallest distance travelled upwards.

Solution.

The horizontal take-off velocity \(v\) is given by the formula:

We use the well-known equation \(h = \frac{{g{t^2}}}{2}\) to find the airborne time. Then

Hence, the horizontal distance (s) is expressed in the form

Take the derivative:

Find the critical points:

So, the critical points are

Using the First Derivative Test, we can show that the range \(s\) has a maximum at \(h = \frac{H}{2}.\)

Solution.

First we determine the angle \(\alpha\) that gives the maximum range \(L\) for given initial speed.

The initial speed is written in vector form as

where

The \(y-\)coordinate of the cork is given by

The time of flight of the cork is determined from the condition \(y = 0.\) Hence,

Then the range \(L\) is determined by the formula:

The range \(L\) has a maximum value when

Thus, the maximum possible range is equal to

It follows from this equation that

Solution.

Let the distance \({O_1}{O_2}\) between the centers of the spheres be equal to \(L,\) and the light source is at a distance \(x\) from the center of the first sphere (Figure \(10\)).

The illuminated areas are the surface areas of spherical sectors. Their total area is given by the formula

where \({h_1},\) \({h_2}\) are the heights of the corresponding spherical caps.

Given the similarity of the right angled triangles \({AL{O_1}}\) and \({{O_1}AI},\) we can write:

Hence,

A similar expression for the second sphere is as follows:

So, the total illuminated area is

Find the derivative:

By equating the derivative to zero, we obtain the following solution

When passing through zero the derivative changes sign from plus to minus, that is the above point is the maximum of the function \(S\left( x \right).\)

Thus, the illuminated area is the largest when the light source is located at a point, at which the ratio of the squares of the distances from the source to the centers of the spheres is as the ratio of their radii cubed.

Solution.

Suppose that the particle \(P\) has a charge \(q.\) Given that the Coulomb force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between the charges, we can write:

The total force is given by

Take the derivative:

Determine the point where the derivative is zero:

Using the First Derivative Test, one can show that the point \(x = \frac{d}{{1 + \sqrt[3]{{\frac{{{q_B}}}{{{q_A}}}}}}}\) is a point of minimum.

Solution.

Consider an infinitesimal piece of the ring with charge \(dQ.\) The magnitude of the force at a point \(x\) due to the charge is given by Coulomb's law:

The projection of the elementary force along the \(x\)-axis is written in the form

Integrating over the ring, one can show that the magnitude of the total force acting on the charge \(q\) at the point \(x\) is equal to

The derivative of the function \(F\left( x \right)\) is given by

The derivative is equal to zero at the following point:

Using the First Derivative Test, one can show that the point \(x = \frac{R}{{\sqrt 2 }}\) corresponds to the maximum of the electrostatic force.

Solution.

Suppose that the liquid flow rate through the opening is described by Torricelli's Law:

where \(x\) is the height of the opening over the horizontal surface, \(h - x\) is the height of the liquid column above the opening, \(g\) is gravitational acceleration (Figure \(13\)).

Calculate the range of the jet \(y\) assuming that the fluid motion is described by the same equations as the motion of a material point.

In this case, the time for which the liquid flowing out of the opening reaches the ground surface is given by

and the jet range is

Substitute the expression for the velocity \(v\) from Torricelli's Law:

We have obtained an expression for the jet range \(y\) depending on the height of the opening \(x.\) We investigate the function for extreme points:

It is clear that this point is a maximum, because when passing through this value the derivative changes sign from plus to minus.

Thus, the opening for the flowing liquid must be positioned exactly in the middle of the liquid column. In this case, the range of the jet will be the greatest and equal to the height of the liquid column:

(Nice answer, isn't it?).

Solution.

We write the ideal gas equation for \(1\) mole is written in the form

where \(R\) is the universal gas constant.

Substituting the expression for the pressure \(p,\) we have

Calculate the derivative of \(T\left( V \right)\) and equate it to zero:

Using the First Derivative Test, we can see from the sign of \(T^\prime\left( V \right)\) on each interval that the point \(V = \frac{1}{\beta }\) is a point of maximum.

Now we can compute the maximum temperature of the gas:

Solution.

Let the radius of the lake be \(R.\) Consider an arbitrary combined route when a part of the path runs across the lake and the other part along the coast. The route of movement is determined by the inscribed angle \(\alpha = \angle CAB\) (Figure \(14\)).

The central angle \(\angle COB\) is equal to \(2\alpha.\) The length of the portion of the path across the lake is

and the length of the portion of the path along the coast is

where the angle \(\alpha\) is measured in radians and varies in the range \(0 \le \alpha \le {\frac{\pi }{2}}.\) Then the total time of motion is written as follows:

Differentiating, we find:

The derivative is equal to zero when

We investigate the type of extremum using the second derivative:

Since the second derivative is always negative for \(0 \le \alpha \le {\frac{\pi }{2}},\) the above point is a maximum of the function \(t\left( \alpha \right),\) which does not fit for us. It is clear that the function \(t\left( \alpha \right)\) is convex upward in the interval \(\left( {0,{\frac{\pi }{2}}} \right)\) and, hence, reaches its lower value in one of the endpoints of the interval. Calculate the values of \(t\left( \alpha \right)\) at \(\alpha = 0\) and \({\alpha = {\frac{\pi }{2}}}:\)

The case \(\alpha = 0\) corresponds to the motion on the water along the straight line \(AOB,\) and the case \({\alpha = \frac{\pi }{2}}\) describes the movement on foot along the semicircle \(ACB.\) Which of these options has the least time? This depends on the relationship between the speeds \(u\) and \(v.\) If

then boating (with the speed \(u\)) will take less time than walking on the ground (with the speed \(v\text{).}\) And conversely, if

the movement along the coastline will be faster.

In case of equality of these ratios, both routes in terms of elapsed time will be equivalent.

Solution.

The total time \(T\) (in hours) is given by the equation

Let the distance \({\left| {SM} \right|}\) be equal to \(x.\) Using the Pythagorean theorem, we have

Then

Take the derivative:

Determine the critical points:

Using the First Derivative Test, we can show that \(x = 3\text{ km}\) is a point of minimum.

Solution.

The coordinates of car \(A\) are given by

Similarly, for car \(B:\)

The distance \(d\) between the cars is defined by the formula

Substituting the coordinates, we have

The derivative of the function \(d\left( t \right)\) has the form:

The derivative is equal to zero at the following point:

Using the First Derivative Test, one can show that this point corresponds to the minimum of the distance between the cars.

Substituting \(b = 100\) miles and \(v = 50\) miles per hour, we get the particular value of \(t\) (hours):

Solution.

Consider the \(xy\)-plane (Figure \(17\)).

The car \(A\) moves along he positive \(y\)-axis. Its coordinates at any time \(t\) are \(\left( {0,ut} \right).\) The car \(B\) moves along the negative \(x\)-axis and its coordinates at time \(t\) are, respectively, \(\left( {b- vt,0} \right).\)

The distance between the cars is calculated by the Pythagorean theorem:

Consider the extreme values of the function \(d\left( t \right).\) Compute the derivative:

The derivative is equal to zero when

The denominator of the derivative is twice the distance \(d\left( t \right)\) and is always positive. The numerator and hence the entire derivative changes sign from minus to plus when passing through this value. Consequently, the obtained value of \(t\) is a point of minimum. At this point, the distance between the cars will be the smallest. It is determined by the following formula:

Solution.

The strength of the beam \(S\) is given by the formula

where \(k\) is a constant of proportionality.

The constraint equation follows from the Pythagorean theorem:

or

We can substitute \({d^2} = 4{R^2} - {w^2}\) into the objective function \(S:\)

Now the strength of the beam \(S\) (our objective function) is written as a function of the width \(w.\)

Differentiate it to find the critical points:

We can use the Second Derivative Test:

As \(S^{\prime\prime}\left( w \right)\) is always negative, we deal with a point of maximum.

Determine the depth \(d\) of the strongest beam:

So, the strongest beam has the following dimensions: