Solved on Feb 05, 2024

Find two positive numbers with given sum that maximize their product. 59. Sum is 110. 60. Sum is 66.

STEP 1

Assumptions
1. We are looking for two positive real numbers.
2. The sum of these two numbers is given.
3. We need to maximize the product of these two numbers.

STEP 2

Let the two positive real numbers be $x$ and $y$ .

STEP 3

According to the problem, we have the sum of $x$ and $y$ as a constant value. For problem 59, the sum is 110, and for problem 60, the sum is 66.
$x + y = S$
where $S$ is the given sum.

STEP 4

Express $y$ in terms of $x$ and $S$ .
$y = S - x$

STEP 5

The product $P$ of the two numbers $x$ and $y$ is given by:
$P = x \cdot y$

STEP 6

Substitute the expression for $y$ from STEP_4 into the product $P$ .
$P = x \cdot (S - x)$

STEP 7

Expand the product to form a quadratic expression.
$P = Sx - x^2$

STEP 8

To maximize the product $P$ , we need to find the vertex of the parabola represented by the quadratic equation $P = Sx - x^2$ .

STEP 9

The vertex form of a parabola is given by $P = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

STEP 10

The vertex of the parabola in the context of our problem occurs at the axis of symmetry, which is given by $x = -\frac{b}{2a}$ for a quadratic equation $ax^2 + bx + c$ .

STEP 11

Identify the coefficients $a$ and $b$ from the quadratic expression $P = -x^2 + Sx$ .
$a = -1, \quad b = S$

STEP 12

Calculate the $x$ -coordinate of the vertex using the formula from STEP_10.
$x = -\frac{b}{2a} = -\frac{S}{2(-1)} = \frac{S}{2}$

STEP 13

Since the parabola opens downwards (because $a$ is negative), the vertex represents the maximum value of the product $P$ .

STEP 14

Substitute $x = \frac{S}{2}$ back into the equation $y = S - x$ to find the corresponding value of $y$ .
$y = S - \frac{S}{2} = \frac{S}{2}$

STEP 15

We have found that the product is maximized when $x = y = \frac{S}{2}$ .

STEP 16

Now, we will apply the results to the specific sums given in the problems.
For problem 59, where $S = 110$ :

STEP 17

Substitute $S = 110$ into $x = \frac{S}{2}$ .
$x = \frac{110}{2} = 55$

STEP 18

Since $x = y$ , we also have $y = 55$ .
The two positive real numbers with a sum of 110 whose product is a maximum are 55 and 55.
For problem 60, where $S = 66$ :

STEP 19

Substitute $S = 66$ into $x = \frac{S}{2}$ .
$x = \frac{66}{2} = 33$

STEP 20

Since $x = y$ , we also have $y = 33$ .
The two positive real numbers with a sum of 66 whose product is a maximum are 33 and 33.

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Find two positive numbers with given sum that maximize their product. 59. Sum is 110. 60. Sum is 66.

STEP 1

Assumptions1. We are looking for two positive real numbers.2. The sum of these two numbers is given.3. We need to maximize the product of these two numbers.

STEP 2

Let the two positive real numbers be xxx and yyy.

STEP 3

According to the problem, we have the sum of xxx and yyy as a constant value. For problem 59, the sum is 110, and for problem 60, the sum is 66.x+y=Sx + y = Sx+y=Swhere SSS is the given sum.

STEP 4

Express yyy in terms of xxx and SSS.y=S−xy = S - xy=S−x

STEP 5

The product PPP of the two numbers xxx and yyy is given by:P=x⋅yP = x \cdot yP=x⋅y

STEP 6

Substitute the expression for yyy from STEP_4 into the product PPP.P=x⋅(S−x)P = x \cdot (S - x)P=x⋅(S−x)

STEP 7

Expand the product to form a quadratic expression.P=Sx−x2P = Sx - x^2P=Sx−x2

STEP 8

To maximize the product PPP, we need to find the vertex of the parabola represented by the quadratic equation P=Sx−x2P = Sx - x^2P=Sx−x2.

STEP 9

The vertex form of a parabola is given by P=a(x−h)2+kP = a(x - h)^2 + kP=a(x−h)2+k, where (h,k)(h, k)(h,k) is the vertex of the parabola.

STEP 10

The vertex of the parabola in the context of our problem occurs at the axis of symmetry, which is given by x=−b2ax = -\frac{b}{2a}x=−2ab​ for a quadratic equation ax2+bx+cax^2 + bx + cax2+bx+c.

STEP 11

Identify the coefficients aaa and bbb from the quadratic expression P=−x2+SxP = -x^2 + SxP=−x2+Sx.a=−1,b=Sa = -1, \quad b = Sa=−1,b=S

STEP 12

Calculate the xxx-coordinate of the vertex using the formula from STEP_10.x=−b2a=−S2(−1)=S2x = -\frac{b}{2a} = -\frac{S}{2(-1)} = \frac{S}{2}x=−2ab​=−2(−1)S​=2S​

STEP 13

Since the parabola opens downwards (because aaa is negative), the vertex represents the maximum value of the product PPP.

STEP 14

Substitute x=S2x = \frac{S}{2}x=2S​ back into the equation y=S−xy = S - xy=S−x to find the corresponding value of yyy.y=S−S2=S2y = S - \frac{S}{2} = \frac{S}{2}y=S−2S​=2S​

STEP 15

We have found that the product is maximized when x=y=S2x = y = \frac{S}{2}x=y=2S​.

STEP 16

Now, we will apply the results to the specific sums given in the problems.For problem 59, where S=110S = 110S=110:

STEP 17

Substitute S=110S = 110S=110 into x=S2x = \frac{S}{2}x=2S​.x=1102=55x = \frac{110}{2} = 55x=2110​=55

STEP 18

Since x=yx = yx=y, we also have y=55y = 55y=55.The two positive real numbers with a sum of 110 whose product is a maximum are 55 and 55.For problem 60, where S=66S = 66S=66:

STEP 19

Substitute S=66S = 66S=66 into x=S2x = \frac{S}{2}x=2S​.x=662=33x = \frac{66}{2} = 33x=266​=33

STEP 20

Since x=yx = yx=y, we also have y=33y = 33y=33.The two positive real numbers with a sum of 66 whose product is a maximum are 33 and 33.

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Assumptions
1. We are looking for two positive real numbers.
2. The sum of these two numbers is given.
3. We need to maximize the product of these two numbers.

Let the two positive real numbers be $x$ and $y$ .

According to the problem, we have the sum of $x$ and $y$ as a constant value. For problem 59, the sum is 110, and for problem 60, the sum is 66.
$x + y = S$
where $S$ is the given sum.

Express $y$ in terms of $x$ and $S$ .
$y = S - x$

The product $P$ of the two numbers $x$ and $y$ is given by:
$P = x \cdot y$

Substitute the expression for $y$ from STEP_4 into the product $P$ .
$P = x \cdot (S - x)$

Expand the product to form a quadratic expression.
$P = Sx - x^2$

To maximize the product $P$ , we need to find the vertex of the parabola represented by the quadratic equation $P = Sx - x^2$ .

The vertex form of a parabola is given by $P = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

The vertex of the parabola in the context of our problem occurs at the axis of symmetry, which is given by $x = -\frac{b}{2a}$ for a quadratic equation $ax^2 + bx + c$ .

Identify the coefficients $a$ and $b$ from the quadratic expression $P = -x^2 + Sx$ .
$a = -1, \quad b = S$

Calculate the $x$ -coordinate of the vertex using the formula from STEP_10.
$x = -\frac{b}{2a} = -\frac{S}{2(-1)} = \frac{S}{2}$

Since the parabola opens downwards (because $a$ is negative), the vertex represents the maximum value of the product $P$ .

Substitute $x = \frac{S}{2}$ back into the equation $y = S - x$ to find the corresponding value of $y$ .
$y = S - \frac{S}{2} = \frac{S}{2}$

We have found that the product is maximized when $x = y = \frac{S}{2}$ .

Now, we will apply the results to the specific sums given in the problems.
For problem 59, where $S = 110$ :

Substitute $S = 110$ into $x = \frac{S}{2}$ .
$x = \frac{110}{2} = 55$

Since $x = y$ , we also have $y = 55$ .
The two positive real numbers with a sum of 110 whose product is a maximum are 55 and 55.
For problem 60, where $S = 66$ :

Substitute $S = 66$ into $x = \frac{S}{2}$ .
$x = \frac{66}{2} = 33$

Since $x = y$ , we also have $y = 33$ .
The two positive real numbers with a sum of 66 whose product is a maximum are 33 and 33.