# Calculus

## Applications of the Derivative # 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 $$F\left( {x,y} \right)$$ 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.

$x = y = \sqrt a .$

### 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 $$x = 16,$$ $$y = 16.$$