Respuesta :
Answer:
20 ft by 10 ft.
Step-by-step explanation:
Let the dimension of the rectangular enclosure be x ft by y ft
- Area to be enclosed [tex]=200$ ft^2[/tex]
- Therefore: Area=xy=200
Perimeter, P(x,y)=2(x+y)=x+x+2y
Fencing material along three sides costs $5 per foot
Fencing material for the fourth side costs $15 per foot.
Therefore, Cost of Fencing=$(15x+5x+5y+5y)
C(x,y)=20x+10y
We can write the cost function as a function of one variable by substituting for y.
Recall: xy=200
[tex]y=\dfrac{200}{x}[/tex]
Therefore:
[tex]C(x)=20x+10(\dfrac{200}{x})\\C(x)=\dfrac{20x^2+2000}{x}[/tex]
To determine dimensions of the enclosure that is most economical to construct, we minimize C(x) by taking its derivative and solving for its critical points.
[tex]C'(x)=\dfrac{20x^2-2000}{x^2}\\$When C'(x)=0\\20x^2-2000=0\\20x^2=2000\\x^2=100\\x^2=10^2\\x=10$ ft[/tex]
Recall that:
[tex]y=\dfrac{200}{x}\\y=\dfrac{200}{10}\\y=20$ ft[/tex]
Therefore, the dimension of the enclosure that is most economical to construct is 20 ft by 10 ft.
- Where the side length of 20 ft is made of materials that costs $5 per foot
- One of the side of length 10 ft is made of materials that costs $5 per foot
- The other side of length 10 ft is made of materials that costs $15 per foot
