Respuesta :
For these questions, we need to form a general equation that satisfies the conditions given.
We know:
A = 320ft²
3 fences - 5 dollars per foot and 1 fence - 12 dollars per foot.
A = 320 = xy
Eliminate one variable and solve for that:
x = [tex] \frac{320}{y} [/tex] (1)
We need to now find an equation for the amount of money we are going to spend.
We can visualise it in our heads: We have two short sides to be built and two long sides to be built. One side will cost us 12 dollars, while the other three costs just 5 dollars to build.
We can now find our perimeter, which is given by 2x + 2y, but we know that x can be replaced as [tex] \frac{320}{y} [/tex].
Hence, our perimeter becomes: 2([tex] \frac{320}{y} [/tex]) + 2y.
Now, let's write an equation for C(y), where C is our total cost.
C(y) = 5*(2*[tex] \frac{320}{y} [/tex] + y) + 12y) = 10*[tex] \frac{320}{y} [/tex] + 5y + 12y = [tex] \frac{3200}{y} [/tex] + 17y
Now, we want to find when it's a minimum so that it will cost us the least amount of money.
Take the first derivative, and let it equal to 0.
C'(y) = -3200*y⁻² + 17 = 0
[tex] \frac{-3200}{y²} [/tex] = -17
[tex] \frac{3200}{y²} [/tex] = 17
3200 = 17y²
y² = [tex] \frac{3200}{17} [/tex]
y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex]
Now, to prove that it's a min. at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], take the second derivative:
C''(y) = 3200y⁻³ and at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], C''(y) > 0 and therefore, min. amount used.
So, at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], it will be the least amount of money spent.
Substitute y, back into (1) to find for x.
x = [tex] \frac{320}{\frac{\sqrt{3200}}{\sqrt{17}}} [/tex]
therefore, x = [tex] \frac{320}{\frac{\sqrt{3200}}{\sqrt{17}}} [/tex] and y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex]
We know:
A = 320ft²
3 fences - 5 dollars per foot and 1 fence - 12 dollars per foot.
A = 320 = xy
Eliminate one variable and solve for that:
x = [tex] \frac{320}{y} [/tex] (1)
We need to now find an equation for the amount of money we are going to spend.
We can visualise it in our heads: We have two short sides to be built and two long sides to be built. One side will cost us 12 dollars, while the other three costs just 5 dollars to build.
We can now find our perimeter, which is given by 2x + 2y, but we know that x can be replaced as [tex] \frac{320}{y} [/tex].
Hence, our perimeter becomes: 2([tex] \frac{320}{y} [/tex]) + 2y.
Now, let's write an equation for C(y), where C is our total cost.
C(y) = 5*(2*[tex] \frac{320}{y} [/tex] + y) + 12y) = 10*[tex] \frac{320}{y} [/tex] + 5y + 12y = [tex] \frac{3200}{y} [/tex] + 17y
Now, we want to find when it's a minimum so that it will cost us the least amount of money.
Take the first derivative, and let it equal to 0.
C'(y) = -3200*y⁻² + 17 = 0
[tex] \frac{-3200}{y²} [/tex] = -17
[tex] \frac{3200}{y²} [/tex] = 17
3200 = 17y²
y² = [tex] \frac{3200}{17} [/tex]
y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex]
Now, to prove that it's a min. at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], take the second derivative:
C''(y) = 3200y⁻³ and at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], C''(y) > 0 and therefore, min. amount used.
So, at y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex], it will be the least amount of money spent.
Substitute y, back into (1) to find for x.
x = [tex] \frac{320}{\frac{\sqrt{3200}}{\sqrt{17}}} [/tex]
therefore, x = [tex] \frac{320}{\frac{\sqrt{3200}}{\sqrt{17}}} [/tex] and y = [tex] \frac{\sqrt{3200}}{\sqrt{17}} [/tex]
