CALCU-Find the length and width of a rectangle tha

Find the length and width of a rectangle that has perimeter 64 meters and a maximum area.

Solution by Steps

step 1

Let the length of the rectangle be

l

and the width be

w

. The perimeter of the rectangle is given by

2l + 2w = 64

step 2

Simplify the perimeter equation:

l + w = 32

step 3

To find the maximum area, we need to express the area

A

in terms of one variable. The area of the rectangle is

A = l \cdot w

step 4

From the perimeter equation, solve for

w

:

w = 32 - l

step 5

Substitute

w

into the area formula:

A = l \cdot (32 - l)

step 6

Simplify the area formula:

A = 32l - l^2

step 7

To find the maximum area, take the derivative of

A

with respect to

l

:

\frac{dA}{dl} = 32 - 2l

step 8

Set the derivative equal to zero to find the critical points:

32 - 2l = 0

step 9

Solve for

l

:

l = 16

step 10

Substitute

l = 16

back into the equation

w = 32 - l

:

w = 32 - 16 = 16

Answer

The length and width of the rectangle that has a perimeter of 64 meters and a maximum area are both 16 meters.

Key Concept

Optimization of area under a constraint

Explanation

To maximize the area of a rectangle with a given perimeter, the rectangle must be a square. In this case, both the length and width are 16 meters.

Find the point on the graph of the function f(x) = sqrt(x) that is closest to the point (18, 0)

Solution by Steps

step 1

Let the point on the graph of the function be

(x, \sqrt{x})

. We need to minimize the distance between this point and

(18, 0)

. The distance formula is given by

d = \sqrt{(x - 18)^2 + (\sqrt{x} - 0)^2}

step 2

To minimize the distance, we can minimize the square of the distance to avoid dealing with the square root. So, we minimize

d^2 = (x - 18)^2 + (\sqrt{x})^2

step 3

Simplify the expression:

d^2 = (x - 18)^2 + x

step 4

Take the derivative of

d^2

with respect to

x

:

\frac{d}{dx}[(x - 18)^2 + x] = 2(x - 18) + 1

step 5

Set the derivative equal to zero to find the critical points:

2(x - 18) + 1 = 0

step 6

Solve for

x

:

2x - 36 + 1 = 0 \implies 2x - 35 = 0 \implies x = 17.5

step 7

The corresponding

y

-coordinate is

\sqrt{17.5}

step 8

Therefore, the point on the graph of

f(x) = \sqrt{x}

that is closest to

(18, 0)

is

(17.5, \sqrt{17.5})

Answer

The point on the graph of

f(x) = \sqrt{x}

that is closest to

(18, 0)

is

(17.5, \sqrt{17.5})

.

Key Concept

Minimizing Distance

Explanation

To find the point on the graph closest to a given point, we minimize the distance between the points by using the distance formula and finding the critical points of the resulting function.

Find two positive numbers whose product is 193 and whose sum is a minimum.

Solution by Steps

step 1

Let the two positive numbers be

x

and

y

. We know that their product is

193

, so we have the equation

xy = 193

step 2

We need to minimize the sum of these two numbers,

S = x + y

step 3

From the product equation

xy = 193

, we can express

y

in terms of

x

:

y = \frac{193}{x}

step 4

Substitute

y

into the sum equation:

S = x + \frac{193}{x}

step 5

To find the minimum sum, we need to take the derivative of

S

with respect to

x

and set it to zero:

\frac{dS}{dx} = 1 - \frac{193}{x^2} = 0

step 6

Solve for

x

:

1 = \frac{193}{x^2} \implies x^2 = 193 \implies x = \sqrt{193}

step 7

Since

x

and

y

are symmetric in the product equation,

y

will also be

\sqrt{193}

step 8

Therefore, the two positive numbers are both

\sqrt{193}

Answer

The two positive numbers are both

\sqrt{193}

.

Key Concept

Optimization of a function

Explanation

To find two numbers whose product is fixed and whose sum is minimized, we express one variable in terms of the other, substitute into the sum equation, and then use calculus to find the minimum value.

Find two positive numbers such that the sum of the first and twice the second is 88 and whose product is a maximum.

Generated Graph

Solution by Steps

step 2

We need to express

y

in terms of

x

. From the equation

x + 2y = 88

, we get

y = \frac{88 - x}{2}

step 3

The product of the two numbers is

P = x \cdot y = x \cdot \frac{88 - x}{2} = \frac{88x - x^2}{2}

step 4

To maximize the product

P

, we need to find the critical points by taking the derivative of

P

with respect to

x

and setting it to zero

step 5

The derivative is

\frac{dP}{dx} = \frac{88 - 2x}{2}

. Setting this equal to zero gives

88 - 2x = 0

, so

x = 44

step 6

Substituting

x = 44

back into the equation

y = \frac{88 - x}{2}

, we get

y = \frac{88 - 44}{2} = 22

step 7

Therefore, the two positive numbers are

x = 44

and

y = 22

A

Key Concept

Optimization

Explanation

The problem involves finding the maximum product of two numbers under a given constraint, which is a typical optimization problem. By expressing one variable in terms of the other and using calculus to find the critical points, we can determine the values that maximize the product.

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