Respuesta :
Answer:
The answer to this question can be described as follows:
Step-by-step explanation:
A)
The goal is to increase the area and therefore
Target formula: A = xy
The limited amount of money, which we have for the fence, and it only has $400
The equation of constraints: 400=(fence costs parallel to roads) +( 3 other side cost)
[tex]\to 400 = 15x+ 5(5y+x)\\\\ \to400 =15x+25y+5x\\\\ \to 400=20x+25y\\[/tex]
B)
To solve y we use the limiting equation:
[tex]\to 400=20x+25y\\\\ \to 25y= 400-20x\\\\\to y=\frac{(400-20x))}{25}\\\\\to y=16-\frac{4}{5}x\\\\[/tex]
Now, put the value in the y and maximise as much as possible into our goal equation:
A function for:
[tex]\to A(x)= x(16-(\frac{4}{5}x)\\\\\to A(x) =16x-\frac{4x^2}{5}[/tex]
C)
Select the critical value the A(x):
[tex]\to \frac{dA}{dx} 16-\frac{8x}{5}=0 \\\\\to 16=\frac{8x}{5}\\\\\to 16 \times 5 =8x\\\\\to 80=8x \\\\\to x=\frac{80}{8}\\\\\to x=10\\\\[/tex]
We now need, with the second derivative, to ensure that this value is maximum:
[tex]\to \frac{d^2A}{dx^2} = \frac{-8}{5}\\\\ \to \frac{d^2A}{dx^2}(10) < O[/tex]
not only is there a relative maximum at x = 10, but we can conclude that the
maximum occurs as[tex]x = 10 \ since \ \frac{d^2A}{dx^2}< O[/tex] for all x. We also need y:
[tex]\to y= 16-\frac{4}{5}x\\\\\to y=16-\frac{4}{5}*10\\\\\to y= 16-8\\\\\to y=8\\\\[/tex]
