Chapter 4.4, Problem 1E

(a.)

To determine

The two numbers.

Answer to Problem 1E

It has been determined that the non-negative numbers are $10$ and $10$ .

Explanation of Solution

Given:

The sum of two non-negative numbers is $20$ .

The sum of their squares is as large as possible; as small as possible.

Concept used:

A function assumes the smallest possible value or the largest possible value at point(s) where its first derivative is zero.

Calculation:

It is given that the sum of two non-negative numbers is $20$ .

Then, the two non-negative numbers can be assumed to be $x$ and $20-x$ .

The sum of their squares is given as,

  $S=x^2+{\left(20-x\right)}^2$

Differentiating,

  $S^{\prime }=2x+2\left(20-x\right)\left(-1\right)$

Simplifying,

  $S^{\prime }=4x-40$

It is given that the sum of their squares is as large as possible; as small as possible.

Then, $S$ is as large as possible; as small as possible.

As discussed, the above implies that $S^{\prime }=0$ .

Put $S^{\prime }=0$ in $S^{\prime }=4x-40$ to get,

  $4x-40=0$

Simplifying,

  $x=10$

Put $x=10$ in $20-x$ to get,

  $20-x=10$

So, the required numbers are $10$ and $10$ .

Conclusion:

It has been determined that the non-negative numbers are $10$ and $10$ .

(b.)

To determine

The two numbers.

Answer to Problem 1E

It has been determined that the required non-negative numbers are $\frac{79}{4}$ and $\frac{1}{4}$ respectively.

Explanation of Solution

Given:

The sum of two non-negative numbers is $20$ .

The sum of one number and the square root of the other number is as large as possible; as small as possible.

Concept used:

A function assumes the smallest possible value or the largest possible value at point(s) where its first derivative is zero.

Calculation:

It is given that the sum of two non-negative numbers is $20$ .

Then, the two non-negative numbers can be assumed to be $20-x$ and $x$ .

The sum of one number and the square root of the other number is given as,

  $S=20-x+\sqrt{x}$

Differentiating,

  $S^{\prime }=-1+\frac{1}{2\sqrt{x}}$

It is given that the sum of one number and the square root of the other number is as large as possible; as small as possible.

Then, $S$ is as large as possible; as small as possible.

As discussed, the above implies that $S^{\prime }=0$ .

Put $S^{\prime }=0$ in $S^{\prime }=-1+\frac{1}{2\sqrt{x}}$ to get,

  $-1+\frac{1}{2\sqrt{x}}=0$

Simplifying,

  $\frac{1}{2\sqrt{x}}=1$

On further simplification,

  $\sqrt{x}=\frac{1}{2}$

Squaring both sides,

  $x=\frac{1}{4}$

Put $x=\frac{1}{4}$ in $20-x$ to get,

  $20-x=\frac{79}{4}$

So, the required numbers are $\frac{79}{4}$ and $\frac{1}{4}$ respectively.

Conclusion:

It has been determined that the required non-negative numbers are $\frac{79}{4}$ and $\frac{1}{4}$ respectively.