# Optimization Problems Involving Numbers

## Solved Problems

Click or tap a problem to see the solution.

### Example 7

Find two positive numbers whose sum is 12 so that the product of the square of one and 4th 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 $$x = 16,$$ $$y = 16.$$