A rectangular dog pen is constructed using a barn wall as one side and 64 m of fencing for the other three sides what is the maximum area of the dog pen

The answer is below

Step-by-step explanation:

Let x represent the length of the rectangular pen and y represent the width of the rectangular pen.

One side of the pen uses the barn wall, hence the remaining side is 64 m, hence:

x + 2y = 64

x = 64 - 2y       (1)

The area of the wall is the product of the length and the width, hence:

Area (A) = length * width = xy = (64 - 2y)y = 64y - 2y²

A = 64y - 2y²

The maximum area of the dog house is gotten when the derivative of the area is equal to 0. Hence:

A' = 64 - 4y = 0

64 - 4y = 0

4y = 64

y = 16 m

Put y = 16, in equation 1 to find x:

x = 64 - 2(16) = 32 m

x = 32 m

Therefore the pen has a length of 32 m and width of 16 m.

A(max) =  512 m²

Dimensions:

x  =  32  m

y  =  16  m

Step-by-step explanation:

Area of the rectangular pen:

A(p) = x*y            x   and   y  sides of the rectangle. A barn will be used in place of a fence; let´s say that only one  x side will be fenced, then the perimeter of the rectangle is:

p  =  x  +  2*y         64  =  x  * 2*y      y  =  (  64  -  x ) / 2

A(p)  =  x*y

Area as a function of x

A(x)  =  x * ( 64  -  x  )/ 2

A(x)  = ( 64*x - x² ) /2

Tacking derivatives on both sides of the equation

A´(x)  = ( 64  -2*x )/2

A´(x)  =  0        64  - 2*x  = 0         2*x  =  64     x  =  32  m

y  =  (  64 - x ) / 2        y  =  (  64  -  32 )/2        y  = 16 m

A(max) =  32 *  16  = 512 m²

To check if x = 32 will bring a maximum to function A(x)

We take the second derivative

A´´(x)  = -  x/4          A´´(x)  < 0   then  x  =  32 is a local maximum

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.