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Sholpan [36]
3 years ago
10

A fence for a rectangular garden with one side against an existing wall is constructed by using 60 feet of fencing. What is the

maximum area that can be enclosed
Mathematics
1 answer:
Y_Kistochka [10]3 years ago
3 0

Step-by-step explanation:

A fence for a rectangular garden with one side against an existing wall is constructed by using 60 feet of fencing.

Perimeter of rectangle (3 sides)= 60 feet

Let 'x' be the width of the wall

Perimeter = 2(length)+2(width)\\60=2(length)+2x\\\\60-2x=2(length)\\\frac{60-2x}{2} =length\\Length =30-x

Formula for the area of the rectangle is

Area=length \cdot width\\A=length(x)

Replace the length we got using perimeter

A=(30-x)(x)\\A(x)= 30x-x^2

To find out the maximum are we take derivative

A'(x)= 30-2x\\0=30-2x\\-30=-2x\\x=15

find out second derivative to check whether x=15 is maximum

A''(x)=-2

second derivative is negative

So, Maximum area at x=15

To find maximum area we plug in 15 for x in A(x)

A(x)=30x-x^2\\A(15)=30(15)-15^2\\A(15)=225

So, maximum area is 225 square feet

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Answer:

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Step-by-step explanation:

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The directix y=5 must intercept the axis of the parabola at the point (3,5), and the vertex is the midpoint between this point and the focus:

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Replacing the values in the equation:

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write a slope intercept equation for a line passing through the point (2,-2) that is parallel and perpendicular to the line x=-1
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You would need two different lines to complete this as lines cannot be both parallel and perpendicular (these are opposites). The answers would be:

Parallel: x = 2

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In order to find these, we first need to see that the original line of x = -1 is a horizontal line. Therefore, any line that is parallel should be horizontal as well. To get a horizontal line through the point (2, -2), the only option is x = 2.

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