![\bf \textit{the area of either pen is 2744, thus}\\\\ A=xy\implies 2744=xy\implies \cfrac{2744}{x}=y](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bthe%20area%20of%20either%20pen%20is%202744%2C%20thus%7D%5C%5C%5C%5C%0AA%3Dxy%5Cimplies%202744%3Dxy%5Cimplies%20%5Ccfrac%7B2744%7D%7Bx%7D%3Dy)
![\bf \begin{array}{llll} \textit{perimeter of the garden pen}\\\\ P=2(x+y)\\\\ P=2\left( x+ \cfrac{2744}{x}\right)\\\\ P=2x+\cfrac{5488}{x}\\ ----------\\ \textit{cost of it}\\\\ C=15\left( 2x+\cfrac{5488}{x} \right)\\\\ C=30x+\cfrac{82320}{x} \end{array}\qquad \begin{array}{llll} \textit{perimeter of the dog pen}\\\\ P=2x+y\\\\ P=2x+\cfrac{2744}{x}\\ ----------\\ \textit{cost of it}\\\\ C=5\left( 2x+\cfrac{2744}{x} \right)\\\\ C=10x+\cfrac{27440}{x} \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Bperimeter%20of%20the%20garden%20pen%7D%5C%5C%5C%5C%0AP%3D2%28x%2By%29%5C%5C%5C%5C%0AP%3D2%5Cleft%28%20x%2B%20%5Ccfrac%7B2744%7D%7Bx%7D%5Cright%29%5C%5C%5C%5C%0AP%3D2x%2B%5Ccfrac%7B5488%7D%7Bx%7D%5C%5C%0A----------%5C%5C%0A%0A%5Ctextit%7Bcost%20of%20it%7D%5C%5C%5C%5C%0AC%3D15%5Cleft%28%202x%2B%5Ccfrac%7B5488%7D%7Bx%7D%20%5Cright%29%5C%5C%5C%5C%0AC%3D30x%2B%5Ccfrac%7B82320%7D%7Bx%7D%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Bperimeter%20of%20the%20dog%20pen%7D%5C%5C%5C%5C%0AP%3D2x%2By%5C%5C%5C%5C%0AP%3D2x%2B%5Ccfrac%7B2744%7D%7Bx%7D%5C%5C%0A----------%5C%5C%0A%5Ctextit%7Bcost%20of%20it%7D%5C%5C%5C%5C%0AC%3D5%5Cleft%28%202x%2B%5Ccfrac%7B2744%7D%7Bx%7D%20%5Cright%29%5C%5C%5C%5C%0AC%3D10x%2B%5Ccfrac%7B27440%7D%7Bx%7D%0A%5Cend%7Barray%7D)
notice... the dog's pen perimeter, does not include the side that's bordering the garden's, since that side will use the heavy duty fence, instead of the light one
so, the sum of both of those costs, will be the C(x)
![\bf C(x)=\left( 30x+\cfrac{82320}{x} \right)+\left( 10x+\cfrac{27440}{x} \right) \\\\\\ C(x)=40x+\cfrac{109760}{x}](https://tex.z-dn.net/?f=%5Cbf%20C%28x%29%3D%5Cleft%28%2030x%2B%5Ccfrac%7B82320%7D%7Bx%7D%20%5Cright%29%2B%5Cleft%28%2010x%2B%5Ccfrac%7B27440%7D%7Bx%7D%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AC%28x%29%3D40x%2B%5Ccfrac%7B109760%7D%7Bx%7D)
so, just take the derivative of it, and set it to 0 to find the extremas, and do a first-derivative test for any minimum
The net cannot be folded to form a pyramid because the faces that are not a base are not all triangles
If you fold this net up, you will get a triangular prism, NOT A PYRAMID.
A pyramid can have ANY polygon as its base, as long as all the other rest of the shapes are triangles.
Depending on the base, the number of triangles in a net of a pyramid must match the number of sides its particular base has.
For example, if you have a square pyramid turned into a net:
The base is a square (4 sides)
There should be 4 triangles on each side.
Because a pyramid is where all the triangles must meet up at a point.
Hope this helps!
X-y=-10
x=-10-y
7x+7y=-14
7(-10-y)+7y=-14
-70-7y+7y=-14
0y=56
y=0
x=-10-y
x=-10-0
x=-10
I’m not 100% sure if this is right, so I hope someone checks this over :)
We will have the following:
* If Kristin does not decrease the price of her cakes, her projected weekly revenue from cake sales will be $2500.
*If Kristin decreases the price of her cakes, her projected weekly revalue will be $2520.
*Kristin will obtain the same revenue if she sells the cakes for $24 or $21.