Calculus

Applications of the Derivative

Optimization Problems in Geometry

Page 1
Problems 1-3
Page 2
Problems 4-15

In geometry, there are many problems in which we want to find the largest or smallest value of a function. As a function, we can consider the perimeter or area of a figure or, for example, the volume of a body. As an independent variable of the function, we can take a parameter of the figure or body such as the length of a side, the angle between two sides, etc. Once the function is composed, it is necessary to investigate it for extreme values using derivatives. It is important to bear in mind that the functions in such problems usually exist on a finite interval, which is determined by the geometry of the system and/or the conditions of the problem.

Solved Problems

Click on problem description to see solution.

 Example 1

The point \(A\left( {a,b} \right)\) is given in the first quadrant of the coordinate plane. Draw a straight line passing through this point, which cuts from the first quadrant the triangle with the smallest area (Figure \(1\)).

 Example 2

An isosceles trapezoid is circumscribed about a circle of radius \(R\) (Figure \(2\)). At what base angle \(\alpha\) the area of the shaded region is the smallest?

 Example 3

A windows has the shape of a rectangle and surmounted by a semicircle (Figure \(3\)). The perimeter of the window is \(P.\) Determine the radius of the semicircle \(R\) that will allow the greatest amount of light to enter.

 Example 4

A rectangle with sides parallel to the coordinate axes and with one side lying along the \(x\)-axis is inscribed in the closed region bounded by the parabola \(y = c – {x^2}\) and the \(x\)-axis. Find the largest possible area of the rectangle.

 Example 5

Two channels of width \(a\) and \(b\) are connected to each other at right angles (Figure \(5\)). Determine the maximum length of logs that can be floated through this system of channels.

 Example 6

A rectangle with sides parallel to the coordinate axes is inscribed in an ellipse (Figure \(6\)) defined by the equation

\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\]

Find the sides of the rectangle with the greatest area.

 Example 7

A picture of height \(a\) is hung on a wall in such a way that its bottom edge is \(h\) units above the eye level. At what distance \(x\) should the observer stand from the wall to be the most favourable for viewing the picture (Fig.\(7a\))?

 Example 8

Two sides of a parallelogram lie on the sides of a triangle, and one vertex of the parallelogram belongs to the third side of the triangle (Figure \(8\)). Find the conditions under which the area of the parallelogram is greatest.

 Example 9

Find the cylinder with the smallest surface area.

 Example 10

Determine the largest volume of a cylinder inscribed in a cone with height \(H\) and base radius \(R\) (Figure \(10a\)).

 Example 11

Find a cone with the largest volume inscribed in a sphere of radius \(R.\)

 Example 12

A body has the shape of a cylinder, the bases of which are surmounted by hemispheres (Figure \(12\)). Determine the height of the cylinder \(H\) and radius of the hemispheres \(R,\) at which the surface area at a fixed volume \(V\) is the smallest.

 Example 13

A cylinder is inscribed in a sphere of radius \(a.\) Find the base radius \(R\) and height \(H\) of the cylinder with the largest possible volume.

 Example 14

Suppose that we have a log of length \(H\) in the form of a frustum of a cone with radii \(R\) and \(r\) (\(R > r\)). From this log, it is required to cut out a beam with the largest possible volume. The beam has the form of a rectangular parallelepiped with a square cross section and the same axis as the log.

 Example 15

A cone has a volume \(V.\) What base radius \(R\) and height \(H\) will minimize the lateral surface area of the cone?

Example 1.

The point \(A\left( {a,b} \right)\) is given in the first quadrant of the coordinate plane. Draw a straight line passing through this point, which cuts from the first quadrant the triangle with the smallest area (Figure \(1\)).

Solution.

Consider the triangles \(OBC\) and \(MBA.\) These triangles are similar. Consequently, the
following relation holds:

\[{\frac{{OC}}{{MA}} = \frac{{OB}}{{MB}}\;\;}\kern0pt{\text{or}\;\;\frac{y}{b} = \frac{x}{{x – a}},}\]

where the coordinates \(x\) and \(y\) satisfy the inequalities \(x > a,\) \(y > b.\) From the last equation we express \(y\) in terms of \(x:\)

\[y = \frac{{bx}}{{x – a}}.\]

The area of the triangle is described by the function \(S\left( x \right):\)

A triangle in the first quadrant with the least area

Figure 1.

\[
{S\left( x \right) = \frac{{xy}}{2} = \frac{x}{2} \cdot \frac{{bx}}{{x – a}} }
= {\frac{{b{x^2}}}{{2\left( {x – a} \right)}}.}
\]

Compute the derivative:

\[
{S’\left( x \right) = {\left( {\frac{{b{x^2}}}{{2\left( {x – a} \right)}}} \right)^\prime } }
= {\frac{b}{2}{\left( {\frac{{{x^2}}}{{x – a}}} \right)^\prime } }
= {\frac{b}{2} \cdot \frac{{2x\left( {x – a} \right) – {x^2}}}{{{{\left( {x – a} \right)}^2}}} }
= {\frac{b}{2} \cdot \frac{{2{x^2} – 2ax – {x^2}}}{{{{\left( {x – a} \right)}^2}}} }
= {\frac{{bx\left( {x – 2a} \right)}}{{{{\left( {x – a} \right)}^2}}}.}
\]

The function \(S\left( x \right)\) has the critical points \(x = 0,\) \(x = a,\) \(x = 2.\) Since \(x > a,\) the solution is the point \(x = 2a.\) When passing through it the derivative changes sign from minus to plus, i.e. \(x = 2a\) is the minimum point of the function \(S\left( x \right).\)

Find the other leg of the triangle:

\[\require{cancel}
y = \frac{{bx}}{{x – a}} = \frac{{b \cdot 2a}}{{2a – a}} = \frac{{2\cancel{a}b}}{\cancel{a}} = 2b.
\]

Thus, the triangle with the smallest area has legs equal to \(2a\) and \(2b.\)

Example 2.

An isosceles trapezoid is circumscribed about a circle of radius \(R\) (Figure \(2\)). At what base angle \(\alpha\) the area of the shaded region is the smallest?

A trapezoid with the least shaded area

Figure 2.

Solution.

The area of an isosceles trapezoid is determined by the formula

\[{S_T} = \frac{{a + b}}{2} \cdot h,\]

where \(a, b\) are the bases of the trapezoid, \(h\) is its height. Obviously, \(h = 2R.\) The area of the circle is \({S_K} = \pi {R^2}.\) Then the area of the shaded region is

\[
{S = {S_T} – {S_K}} = {\frac{{a + b}}{2} \cdot 2R – \pi {R^2} }
= {\left( {a + b} \right)R – \pi {R^2}.}
\]

Since the trapezoid is circumscribed about a circle, the sum of opposite sides is the same, i.e.

\[
{a + b = 2\ell\;\;\text{or}\;\;}\kern0pt
{a + b = 2 \cdot \frac{{2R}}{{\sin \alpha }} }
= {\frac{{4R}}{{\sin \alpha }}.}
\]

Here \(\ell\) indicates the leg of the trapezoid. Substituting \(\left( {a + b} \right)\) in the previous relation, we get

\[
{S = S\left( \alpha \right) = \frac{{4R}}{{\sin \alpha }} \cdot R – \pi {R^2} }
= {{R^2}\left( {\frac{4}{{\sin \alpha }} – \pi } \right).}
\]

Investigate the area \(S\left( \alpha \right)\) for extreme values. Calculate the derivative \(S’\left( \alpha \right):\)

\[
{S’\left( \alpha \right) = {\left[ {{R^2}\left( {\frac{4}{{\sin \alpha }} – \pi } \right)} \right]^\prime } }
= {4{R^2}\left( { – \frac{1}{{{{\sin }^2}\alpha }}} \right) \cdot \cos \alpha }
= { – \frac{{4{R^2}\cos \alpha }}{{{{\sin }^2}\alpha }}.}
\]

It is evident that the derivative is zero provided

\[\cos \alpha = 0,\;\; \Rightarrow \alpha = \frac{\pi }{2},\]

and when passing through this point (with increasing \(\alpha\)) the derivative changes sign from minus to plus. Consequently, \(\alpha = \large\frac{\pi }{2}\normalsize\) is the minimum of the function \(S\left( \alpha \right).\) In this case, the trapezoid becomes a square. The minimum value of the area is determined by the formula

\[{S_{\min }} = {R^2}\left( {4 – \pi } \right).\]

Example 3.

A windows has the shape of a rectangle and surmounted by a semicircle (Figure \(3\)). The perimeter of the window is \(P.\) Determine the radius of the semicircle \(R\) that will allow the greatest amount of light to enter.

Solution.

Obviously, one side of the rectangle is equal to \(2R.\) We denote the other side by \(y.\) The perimeter of the window is given by

\[P = \pi R + 2R + 2y.\]

Hence, we find \(y:\)

\[y = \frac{1}{2}\left[ {P – \left( {\pi + 2} \right)R} \right].\]

The area of the window is as follows:

\[
{S = \frac{{\pi {R^2}}}{2} + 2Ry }
= {\frac{{\pi {R^2}}}{2} + 2R \cdot \frac{1}{2}\left[ {P – \left( {\pi + 2} \right)R} \right] }
= {\frac{{\pi {R^2}}}{2} + PR – \pi {R^2} – 2{R^2} }
= {PR – \frac{{\pi {R^2}}}{2} – 2{R^2}.}
\]
A window with the greatest area

Figure 3.

The resulting expression is a function \(S\left( R \right).\) We investigate its extreme points. Find the derivative:

\[
{S’\left( R \right) = {\left( {PR – \frac{{\pi {R^2}}}{2} – 2{R^2}} \right)^\prime } }
= {P – \pi R – 4R }
= {P – \left( {\pi + 4} \right)R.}
\]

Determine the stationary points:

\[
{S’\left( R \right) = 0,\;\;}\Rightarrow
{P – \left( {\pi + 4} \right)R = 0,\;\;}\Rightarrow
{R = \frac{P}{{\pi + 4}}.}
\]

Since the second derivative is negative:

\[
{S^{\prime\prime}\left( R \right) = {\left[ {P – \left( {\pi + 4} \right)R} \right]^\prime } }
= { – \left( {\pi + 4} \right) \lt 0,}
\]

then this point is a maximum, i.e. the area of the window will be the greatest at this value of \(R\)

The maximum value of the area is equal to

\[
{{S_{\max }} = PR – \frac{{\pi {R^2}}}{2} – 2{R^2} }
= {P\left( {\frac{P}{{\pi + 4}}} \right) – \left( {\frac{\pi }{2} + 2} \right){\left( {\frac{P}{{\pi + 4}}} \right)^2} }
= {\frac{{{P^2}}}{{\pi + 4}} – \frac{{\left( {\cancel{\pi + 4}} \right){P^2}}}{{2{{\left( {\pi + 4} \right)}^{\cancel{2}}}}} }
= {\frac{{2{P^2} – {P^2}}}{{2\left( {\pi + 4} \right)}} }
= {\frac{{{P^2}}}{{2\left( {\pi + 4} \right)}}.}
\]
Page 1
Problems 1-3
Page 2
Problems 4-15