How do you find the dimensions of the largest possible garden if you are given four hundred eighty dollars to fence a rectangular garden and the fencing for the north and south sides of the garden costs $10 per foot and the fencing for the east and west sides costs $15 per foot?

Answer 1

North and South sides should be 12 feet long and the other two sides should be 8 feet long.

The problem can be solved using the calculus of one variable or the calculus of two variables using Lagrange multipliers. The initial analysis is the same either way. Because the question does not specify Lagrange Multipliers, I'll do the single variable solution. (It is accessible to more readers.)

Let #x# = the length in feet of the North and South sides (their cost is #$10# per foot).
Let #y# = the length in feet of the East and West sides (their cost is #$15# per foot)
The area is #A=xy#. We want to make #A# as large as possible.
The total cost of the #x# by #y# rectangle is: #C=10x+15y+10x+15y=20x+30y# .
We have #$480# to spend.
Is is clear that we will not get the largest possible area if we spend less than #$480#. So we can see that, for largest area we'll have: #20x+30y=480#

Single Variable Calculus Solution

We want #x# and #y# to make the product #A=xy# as large as possible. We also know that #20x+30y=480#. Solve for #y# to get:
#y=16-2/3x#.
Now substitute in the area: #A=x(16-2/3x)=16x-2/3x^2#

Maximize as usual:

#A'=16-4/3x=0# at #x=3/4*16=12# Clearly #A'# is never undefined. So, #12# is the only critical number for #A#.
#A''(12)<0# So #A(12)# is the global maximum.

We weren't asked or the area, but the dimensions:

#x=12#, #y=2/3x=8#

North and South sides should be 12 feet long and the other 2 sides should be 8 feet long.

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Answer 2

Let the length of the garden be ( L ) feet and the width be ( W ) feet. The total cost ( C ) of fencing is given by the equation ( C = 10(2L) + 15(2W) ), which simplifies to ( C = 20L + 30W ). We are given that ( C = $480 ), so the equation becomes ( 480 = 20L + 30W ). To find the dimensions of the largest possible garden, we need to maximize the area ( A ) of the garden, which is given by ( A = LW ).

First, solve the cost equation for one variable, such as ( W ), to get ( W = (480 - 20L) / 30 ). Substitute this expression for ( W ) into the area equation to get ( A = L(480 - 20L) / 30 ). Simplify this to ( A = (16L^2 - 80L) / 3 ). To find the maximum area, take the derivative of ( A ) with respect to ( L ), set it equal to zero, and solve for ( L ).

The derivative ( A' ) is ( A' = (32L - 80) / 3 ), and setting ( A' ) equal to zero gives ( 32L - 80 = 0 ), which simplifies to ( L = 2.5 ). Substituting this value back into the equation for ( W ) gives ( W = (480 - 20(2.5)) / 30 ), which simplifies to ( W = 13 ).

Therefore, the dimensions of the largest possible garden are ( L = 2.5 ) feet and ( W = 13 ) feet.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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