Mathematics, 23.03.2020 22:07 mwilliams457

A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his property. His neighbor will pay for one third of the cost of materials on that side. Find the dimensions of the playground that will minimize the homeowner's total cost for materials. Follow the steps:

(a) Let the width to be y and the length (the side along the boundary of his property) to be x, and assume that the material costs $1 per foot. Then the quantity to be minimized is (expressed as a function of both x and y) C= 2y+5/3x . (Use fraction for coefficients.)

(b) The condition that x and y must satisfy is y= 200/x

(c) Using the condition to replace y by x in C , C can then be expressed as a function of x: C(x)= 2(200/x)+x+2/3x

(d) The domain of C is (0 , infty). (Use ``infty'' for [infinity]. )

(e) The only critical number of C in the domain is x= 15.5 . (Keep 1 decimal place (rounded)). We use the Second-Derivative Test to classify the critical number as a relative maximum or minimum, or neither:

At the critical number x= 15.5 , the second derivative C'' (15.5 ) is positive . Therefore at x= 15.5 , the function has a relative minimum .

(f) Finally, plug x= 15.5 into the condition of x and y we obtain y= 12.9

Therefore the length and width of the playground that will minimize the homeowner's total cost for materials are x= 15.5 feet and y= 12.9 feet, with the side along the boundary of his property equals 15.5 feet.

(These are the correct answers I just don't know how to get them)

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Answer from: JEThompson6416

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Answer from: Quest

step-by-step explanation:

the equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 4) and (x₂, y₂ ) = (3, - 5) ← 2 points on the line

m = $\frac{-5-4}{3-0}$ = $\frac{-9}{3}$ = - 3

with m = - 3 and (a, b) = (3, - 5), then

y - (- 5) = - 3(x - 3), that is

y + 5 = - 3(x - 3) → b

Answer from: Quest

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.

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