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 \(F\left( {x,y} \right)\) 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

Click or tap a problem to see the solution.

### Example 1

Find two numbers whose sum is \(36\) if the sum of their squares is to be a minimum.### Example 2

Find two positive numbers whose product is \(a\) such that their sum is minimum.### Example 3

Find two numbers whose difference is \(8\) and whose product is a minimum.### Example 4

Determine two positive numbers whose product is \(4\) such that the sum of their squares is minimum.### Example 5

Find the number whose sum with its reciprocal is a minimum.### Example 6

Find two numbers whose difference is \(6\) such that the sum of their squares is a minimum.### 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 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 \(y = \large{\frac{a}{x}}\normalsize\) 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 \(y = -3.\)

### 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 \(x = 16,\) \(y = 16.\)