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g The fencing of the left border costs $10 per foot, while the fencing of the lower border costs $2 per foot. (No fencing is required along the river.) You want to spend $520 and enclose as much area as possible. What are the dimensions of your garden, and what area does it enclose?

Respuesta :

Answer:

[tex]Area = 1690ft^2\\x = 130ft, y = 26ft[/tex]

Step-by-step explanation:

Given the costs we can form an equation:

[tex] 10y + 2x =520 - Eq(A)[/tex]

and fencing is triangular such that the area enclosed can be written as:

[tex]A = \dfrac{xy}{2} -Eq(B)[/tex]

  • First need to convert the above equation so that it is only in terms of one variable. [either x or y]

To make the equation only in terms of [tex]x[/tex] we can substitute [tex]y[/tex] from Eq(A) i.e, [tex]y =\frac{520-2x}{10}[/tex], to Eq(B)

[tex]A = \dfrac{x}{2} \dfrac{520-2x}{10}[/tex]

simplify

[tex]A = \dfrac{520x - 2x^2}{20}[/tex]

[tex]A = -\dfrac{1}{10}x^2 + 26x [/tex]

  • Now, in order to find the maximum area enclosed we can find [tex]\frac{dA}{dx}[/tex] and equate it zero.

[tex]\dfrac{dA}{dx} = -\dfrac{1}{5}x + 26 [/tex]

[tex]0 = -\dfrac{1}{5}x + 26 [/tex]

[tex]-26 = -\dfrac{1}{5}x[/tex]

[tex]x = 130[/tex]

we have the length of one dimension: specifically, the lower fence will be [tex]x =130ft[/tex]

we can use this value of [tex]x[/tex] to find the corresponding value of [tex]y[/tex]. From Eq(A)

[tex] 10y + 2x =520[/tex]

[tex] 10y +2(130) = 520[/tex]

[tex] y = \dfrac{520 - 2(130)}{10}[/tex]

[tex] y = 26[/tex]

the length of the left fence will be [tex]y =26ft[/tex]

  • The enclosed area by the fence will be

[tex]A = \dfrac{xy}{2}[/tex]

[tex]A = \dfrac{(130)(26)}{2}[/tex]

[tex]A = 1690[/tex]

Hence the maximum area that can enclosed by the fences provided the costs will be [tex]1690ft^2[/tex]

  • You can even check the cost of the dimensions whether they all add up to $520 or not.

Use Eq(A)

[tex] 10y + 2x =520 [/tex]

[tex] 10(26) + 2(130) =520 [/tex]

and indeed it does!

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