Mathematics, 04.07.2019 17:00 ziyonsolange

a three sided fence is to be built next to a straight river which forms the fourth side of a rectangular region. the enclosed area is to equal 3200 ft2. the cost of the fence is $2/ft. this question will have you find the dimensions that minimize the cost of the fence.(a) what quantity needs to be optimized? (b) draw a picture modeling this situation.40(c) find a function for the quantity that is being optimized and find its domain.(d) find the dimensions that minimize the cost of the fence.

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Answer from: Indigosiniy

The area of the three sided fence is the product of its dimensions.

(a) The quantity that needs to be optimized is the cost of the perimeter of the fence

(b) See attachment for the picture of the model

The area is given as:

$\mathbf{A = 3200}$

From the figure, the perimeter is given as:

$\mathbf{P = 2x + y}$

So, the area is:

$\mathbf{A = xy = 3200}$

Make y the subject

$\mathbf{y = \frac{3200}x}$

Substitute $\mathbf{y = \frac{3200}x}$ in $\mathbf{P = 2x + y}$

$\mathbf{P = 2x + \frac{3200}{x}}$

The fence costs $2 per 1 ft.

So, we have:

$\mathbf{P = 2 \times (2x + \frac{3200}{x})}$

$\mathbf{P = 4x + \frac{6400}{x}}$

So, the function to optimize is:

$\mathbf{P(x) = 4x + \frac{6400}{x}}$

And the domain is:

$\mathbf{x 0}$

(d) The dimension that optimizes the fence

We have:

$\mathbf{P(x) = 4x + \frac{6400}{x}}$

Differentiate

$\mathbf{P'(x) = 4 - \frac{6400}{x^2}}\\$

Set to 0

$\mathbf{4 - \frac{6400}{x^2} = 0}$

Rewrite as:

$\mathbf{4 = \frac{6400}{x^2} }$

Cross multiply

$\mathbf{4x^2 = 6400}$

Divide both sides by 4

$\mathbf{x^2 = 1600}$

Take positive square roots

$\mathbf{x = 40}$

Substitute $\mathbf{x = 40}$ in $\mathbf{y = \frac{3200}x}$

$\mathbf{y = \frac{3200}{40}}$

$\mathbf{y = 80}$

Hence, the dimension that minimizes the cost of the fence is 40 ft by 80 ft

Read more about minimum and maximum areas at:

link

a three sided fence is to be built next to a straight river which forms the fourth sid

Answer from: datbih12345

Let x ft be the width of the fence and y ft be the length of the fence.

1. The area of rectangular region is

x·y=3200 sq. ft.

2. The perimeter of the fence is x+y+x = 2x+y ft (fourth side is river). The total cost of the fence is $2(2x+y). This quantity needs to be optimized.

3. Note that x>0 and y>0. See attached diagram for graphs.

5. Since x·y=3200, then

$y=\dfrac{3200}{x}$

and function

$f(x)=2(2x+y)=4x+2y=4x+\dfrac{6400}{x}$

is the function for the quantity that is being optimized.

The domain of this function is $x\neq 0.$

6. Find the derivative f'(x):

$f'(x)=4-\dfrac{6400}{x^2}.$

When f'(x)=0, you have

$4-\dfrac{6400}{x^2}=0,\\ \\4x^2-6400=0,\\ \\x^2-1600=0,\\ \\x^2=1600,\\ \\x_1=40,\ x_2=-40.$

The width of the fence must have positive length, then x=40 ft and y=80 ft.

Answer from: Quest

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine equals adjacent

tangent over hypotenuse

Special right triangles trig ratios and circles need i am studying for my test

Answer from: Quest

answer: 1722

step-by-step explanation:

given: the mass of saturn= $5.685\times10^{26}\ kgs$

the mass of mercury= $3.302\times10^{23}\ kgs$

the number of times the mass of saturn is greater than the mass of mercury

$=\frac{\text{mass of saturn}}{\text{mass of mercury}}\\=\frac{5.685\times10^{26}}{3.302\times10^{23}}\\\\=1.72168\times10^{26-23}{by using law of exponent}]\\=1.72168\times10^3=1721.68\approx1722$

hence, the mass of saturn is 1722 times greater than the mass of mercury.

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