Calculus

Applications of the Derivative

Applications of Derivative Logo

Optimization Problems Involving Numbers

Number problems involve finding two numbers that satisfy certain conditions.

If we label the numbers using the variables x and y, we can compose the objective function F (x, y) to be maximized or minimized.

The constraint specified in the problem allows to eliminate one of the variables.

When we get the objective function as a single variable function, we can use differentiation to find the extreme values.

Solved Problems

Example 1.

Find two numbers whose sum is \(36\) if the sum of their squares is to be a minimum.

Solution.

Let \(x\) and \(y\) be the two numbers. We want to find the minimum of the function

\[{F\left( {x,y} \right)} = {{x^2} + {y^2}}.\]

As \(x + y = 36,\) we can eliminate one variable in the objective function. Substitute \(y = 36 - x\) in the objective function.

\[F\left( {x,y} \right) = {x^2} + {y^2} = {x^2} + {\left( {36 - x} \right)^2} = {x^2} + 1296 - 72x + {x^2} = 2{x^2} - 72x + 1296 \equiv F\left( x \right).\]

Take the derivative:

\[F^\prime\left( x \right) = \left( {2{x^2} - 72x + 1296} \right)^\prime = 4x - 72.\]

The critical points are

\[F^\prime\left( x \right) = 0,\;\; \Rightarrow 4x - 72 = 0,\;\; \Rightarrow x = 18.\]

Note that the second derivative is positive:

\[F^{\prime\prime}\left( x \right) = \left( {4x - 72} \right)^\prime = {4 \gt 0}.\]

Hence, the objective function has a local minimum at \(x = 18.\)

So the sum of squares is a minimum when \(x = 18,\) \(y = 36-x = 18.\)

Example 2.

Find two positive numbers whose product is \(a\) such that their sum is minimum.

Solution.

Let \(x\) and \(y\) be the two numbers. The objective function is written in the form

\[F = x + y.\]

As \(xy = a,\) we can substitute \(y = \frac{a}{x}\) into the objective function:

\[F = x + y = x + \frac{a}{x} = F\left( x \right).\]

Take the derivative and find the critical points:

\[F^\prime\left( x \right) = \left( {x + \frac{a}{x}} \right)^\prime = 1 - \frac{a}{{{x^2}}};\]
\[F^\prime\left( x \right) = 0,\;\; \Rightarrow 1 - \frac{a}{{{x^2}}} = 0,\;\; \Rightarrow {x^2} = a,\;\; \Rightarrow x = \pm \sqrt a .\]

We should take only positive root \(x = + \sqrt a .\)

Find the second derivative and determine its sign at this point:

\[F^{\prime\prime}\left( x \right) = \frac{{3a}}{{{x^3}}},\;\; \Rightarrow F^{\prime\prime}\left( x \right) = \frac{{3a}}{{{{\left( {\sqrt a } \right)}^3}}} = \frac{3}{{\sqrt a }} \gt 0.\]

We see that \(x = \sqrt a\) is a point of minimum by the Second Derivative Test.

Hence, the answer is

\[x = y = \sqrt a .\]

Example 3.

Find two numbers whose difference is \(8\) and whose product is a minimum.

Solution.

The objective function is

\[F = xy,\]

where \(x\) and \(y\) are the two numbers.

Since \(x - y = 8,\) we can substitute \(y = x - 8\) in the objective function above. This yields:

\[F = xy = x\left( {x - 8} \right) = {x^2} - 8x = F\left( x \right).\]

Take the derivative:

\[F^\prime\left( x \right) = \left( {{x^2} - 8x} \right)^\prime = 2x - 8.\]

There is one critical point \(x = 4\).

Note that the second derivative is always positive:

\[F^{\prime\prime}\left( x \right) = \left( {2x - 8} \right)^\prime = 2 \gt 0.\]

Hence, the objective function has a minimum at the point \(x = 4.\) The other number equals \(y = -4.\)

Example 4.

Determine two positive numbers whose product is \(4\) such that the sum of their squares is minimum.

Solution.

The objective function is given by

\[F = {x^2} + {y^2},\]

where \(x,y\) are the two unknown numbers.

As \(xy = 4,\) we obtain:

\[F = {x^2} + {y^2} = {x^2} + {\left( {\frac{4}{x}} \right)^2} = {x^2} + \frac{{16}}{{{x^2}}} = F\left( x \right).\]

The derivative of the objective function is

\[F^\prime\left( x \right) = \left( {{x^2} + \frac{{16}}{{{x^2}}}} \right)^\prime = 2x - \frac{{32}}{{{x^3}}} = \frac{{2{x^4} - 32}}{{{x^3}}}.\]

Now it is easy to find the critical points:

\[F^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{2{x^4} - 32}}{{{x^3}}} = 0,\;\; \Rightarrow {x^4} = 16,\;\; \Rightarrow x = \pm 2,\]

so the critical points are

\[x = - 2,\,0,\,2.\]

We should take only the point \(x = 2.\) Then \(y = 2.\)

Example 5.

Find the number whose sum with its reciprocal is a minimum.

Solution.

The function to be minimized is written as

\[F\left( x \right) = x + \frac{1}{x},\]

where \(x\) is supposed to be a positive number.

Take the derivative:

\[F^\prime\left( x \right) = \left( {x + \frac{1}{x}} \right)^\prime = 1 - \frac{1}{{{x^2}}} = \frac{{{x^2} - 1}}{{{x^2}}}.\]

There are the following critical values:

\[x = - 1,\,0,\,1.\]

Only the root \(x = 1\) satisfies the condition \(x \gt 0.\)

Determine the second derivative:

\[F^{\prime\prime}\left( x \right) = \left( {1 - \frac{1}{{{x^2}}}} \right)^\prime = \frac{2}{{{x^3}}} \gt 0.\]

As the second derivative is positive at \(x = 1,\) this point corresponds to the minimum of the objective function. The minimum value of the function is

\[{F_{\min }} = F\left( 1 \right) = 1 + \frac{1}{1} = 2.\]

Example 6.

Find two numbers whose difference is \(6\) such that the sum of their squares is a minimum.

Solution.

Let \(x\) and \(y\) be the two numbers. The objective function is written as

\[F = {x^2} + {y^2}.\]

As \(x - y = 6,\) we substitute \(y = x - 6\) in the function above:

\[F = {x^2} + {y^2} = {x^2} + {\left( {x - 6} \right)^2} = {x^2} + {x^2} - 12x + 36 = 2{x^2} - 12x + 36 = F\left( x \right).\]

Compute the derivative:

\[F^\prime\left( x \right) = \left( {2{x^2} - 12x + 36} \right)^\prime = 4x - 12,\]

so the critical point is \(x = 3.\)

The second derivative is

\[F^{\prime\prime}\left( x \right) = \left( {4x - 12} \right)^\prime = 4 \gt 0.\]

Hence, \(x = 3\) corresponds to the minimum of the objective function by the Second Derivative Test. The other number equals \(y = -3.\)

See more problems on Page 2.

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