Calculus

Applications of the Derivative

Applications of Derivative Logo

Optimization Problems Involving Numbers

Solved Problems

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Example 7

Find two positive numbers whose sum is \(12\) so that the product of the square of one and \(4\text{th}\) power of the other is maximum.

Example 8

Find two positive numbers whose product is \(2\) and the sum of one number and the square of the other is a minimum.

Example 9

Find two positive numbers whose sum is \(7\) and the product of the cube of one number and the exponential function of the other is a maximum.

Example 10

The sum of two positive numbers is \(24.\) The product of one and the square of the other is maximum. Find the numbers.

Example 11

Find two positive numbers whose sum is \(32\) and the sum of their square roots is maximum.

Example 7.

Find two positive numbers whose sum is \(12\) so that the product of the square of one and \(4\text{th}\) power of the other is maximum.

Solution.

The objective function is written in the form

\[F\left( {x,y} \right) = {x^2}{y^4},\]

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

As \(x + y = 12\) we can write

\[F = {x^2}{y^4} = {x^2}{\left( {12 - x} \right)^4} = F\left( x \right).\]

Compute the derivative:

\[F^\prime\left( x \right) = \left[ {{x^2}{{\left( {12 - x} \right)}^4}} \right]^\prime = 2x \cdot {\left( {12 - x} \right)^4} + {x^2} \cdot 4{\left( {12 - x} \right)^3} \cdot \left( { - 1} \right) = 6x{\left( {12 - x} \right)^3}\left( {4 - x} \right).\]

Determine the critical points:

\[F^\prime\left( x \right) = 0,\;\; \Rightarrow 6x{\left( {12 - x} \right)^3}\left( {4 - x} \right) = 0,\;\; \Rightarrow x = 0,\,4,\,12.\]

At \(x = 0\) and \(x = 12,\) the objective function is equal to zero.

When \(x = 4,\) the value of \(y\) is

\[y = 12 - x = 12 - 4 = 8.\]

At this point, the objective function attains the maximum value:

\[{F_{\max }} = {4^2} \cdot {8^4} = {2^{16}} = 65536.\]

Example 8.

Find two positive numbers whose product is \(2\) and the sum of one number and the square of the other is a minimum.

Solution.

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

\[xy = 2,\;\; \Rightarrow y = \frac{2}{x}.\]

The objective function is given by

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

Find the derivative and determine the critical points:

\[F^\prime\left( x \right) = \left( {x + \frac{4}{{{x^2}}}} \right)^\prime = 1 + 4 \cdot \left( { - \frac{2}{{{x^3}}}} \right) = 1 - \frac{8}{{{x^3}}} = \frac{{{x^3} - 8}}{{{x^3}}};\]
\[F^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{{x^3} - 8}}{{{x^3}}} = 0,\;\; \Rightarrow {x^3} = 8,\;\; \Rightarrow x = 2.\]

Thus, the function has two critical points \(x = 0\) and \(x = 2.\) We should take only positive number \(x = 2.\)

Using the First Derivative Test, one can show that \(x = 2\) is a point of minimum.

The second number is \(y = 1.\)

Example 9.

Find two positive numbers whose sum is \(7\) and the product of the cube of one number and the exponential function of the other is a maximum.

Solution.

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

\[F = {x^3}{e^y}.\]

As \(x + y = 7,\) we substitute \(y = 7 - x\) in the function above.

\[F = {x^3}{e^y} = {x^3}{e^{7 - x}} = F\left( x \right).\]

Differentiate \(F\left( x \right):\)

\[F^\prime\left( x \right) = \left( {{x^3}{e^{7 - x}}} \right)^\prime = \left( {{x^3}} \right)^\prime \cdot {e^{7 - x}} + {x^3} \cdot \left( {{e^{7 - x}}} \right)^\prime = 3{x^2}{e^{7 - x}} + {x^3} \cdot {e^{7 - x}} \cdot \left( { - 1} \right) = 3{x^2}{e^{7 - x}} - {x^3}{e^{7 - x}} = {x^2}{e^{7 - x}}\left( {3 - x} \right).\]

It is clear that the positive critical value is only \(x = 3.\) Using the First Derivative Test, one can show that \(x = 3\) is a point of local maximum.

Respectively, the other number is \(y = 4.\)

Example 10.

The sum of two positive numbers is \(24.\) The product of one and the square of the other is maximum. Find the numbers.

Solution.

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

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

The constraint equation has the form

\[x + y = 24,\;\; \Rightarrow y = 24 - x.\]

Hence

\[F = x{y^2} = x{\left( {24 - x} \right)^2}.\]

Expanding \({\left( {24 - x} \right)^2},\) we obtain:

\[F\left( x \right) = x{\left( {24 - x} \right)^2} = x\left( {576 - 48x + {x^2}} \right) = 576x - 48{x^2} + {x^3}.\]

Differentiate:

\[F^\prime\left( x \right) = \left( {576x - 48{x^2} + {x^3}} \right)^\prime = 576 - 96x + 3{x^2} = 3\left( {192 - 32x + {x^2}} \right).\]

Find the critical points:

\[F^\prime\left( x \right) = 0,\;\; \Rightarrow 3\left( {192 - 32x + {x^2}} \right) = 0,\;\; \Rightarrow {x^2} - 32x + 192 = 0;\]
\[D = {\left( { - 32} \right)^2} - 4 \cdot 192 = 1024 - 768 = 256;\]
\[{x_{1,2}} = \frac{{ - \left( { - 32} \right) \pm \sqrt {256} }}{2} = \frac{{32 \pm 16}}{2} = 24,\,8;\]

When \(x = 24,\) then \(y = 0,\) so the objective function is equal to zero in this case.

Note that the second derivative is

\[F^{\prime\prime}\left( x \right) = \left( {576 - 96x + 3{x^2}} \right)^\prime = 6x - 96.\]

Hence, the second derivative is negative for \(x = 8,\) that is the point \(x = 8\) is a point of maximum of the objective function.

The other number \(y\) is equal to

\[y = 24 - x = 24 - 8 = 16.\]

Example 11.

Find two positive numbers whose sum is \(32\) and the sum of their square roots is maximum.

Solution.

We write the objective function in the form

\[F = \sqrt x + \sqrt y ,\]

where \(x, y\) are two positive numbers.

As \(x + y = 32,\) we can plug in \(y = 32 - x\) into the objective function.

\[F = \sqrt x + \sqrt y = \sqrt x + \sqrt {32 - x} = F\left( x \right).\]

Differentiate \(F\left( x \right):\)

\[F^\prime\left( x \right) = \left( {\sqrt x + \sqrt {32 - x} } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{\left( { - 1} \right)}}{{2\sqrt {32 - x} }} = \frac{{\sqrt {32 - x} - \sqrt x }}{{2\sqrt x \sqrt {32 - x} }}.\]

Determine the critical points:

\[F^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{\sqrt {32 - x} - \sqrt x }}{{2\sqrt x \sqrt {32 - x} }} = 0,\;\; \Rightarrow \sqrt {32 - x} - \sqrt x = 0,\;\; \Rightarrow \sqrt {32 - x} = \sqrt x ,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {32 - x = x}\\ {x \lt 32}\\ {x \gt 0} \end{array}} \right., \Rightarrow x = 16.\]

There are total 3 critical points: \(x = 0, 16, 32.\) We calculate the values of the objective function at these points:

\[F\left( 0 \right) = \sqrt 0 + \sqrt {32 - 0} = \sqrt {32} = 4\sqrt 2 \approx 5.66;\]
\[F\left( {16} \right) = \sqrt {16} + \sqrt {32 - 16} = 4 + 4 = 8;\]
\[F\left( {32} \right) = \sqrt {32} + \sqrt {32 - 32} = \sqrt {32} = 4\sqrt 2 \approx 5.66\]

Thus, the maximum value \({F_{\max }} = 8\) is attained at \(x = 16,\) \(y = 16.\)

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