Question

A pigpen is to be enclosed using 640 m of fencing along three of its sides, the fourth side being a barn. Determine the dimensions of the pen that will maximize the area (100 PTS + BRAINLIEST)

Answer 1

Answer:

$W_{max}=480m\\ L_{max}=160m$

Step-by-step explanation:

Since we know we only have 640 feet of fence available, we know that L + W + L = 640, $\implies$ 2L + W = 640. This allows us to represent the width, W, in terms of L: W = 640 – 2L

Remember, the area of a rectangle is equal to the product of its width and length, therefore,

$A_{\rm pigpen}=L(640-2L)\implies 640L-2L^2$

Notice that, quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area.

recall,

• $x_{max}=-b/2a$
• $f(x)_{max}=f(x_{max})$

Since our function is A(L)=640L-2L², we get

• a=-2
• b=640

plug in the value of a and b into the first formula:

$L_{max}=-(640)/2(-2)\implies 160$

$A(L)_{max}=640(160)-2(160)^2\implies 76800$

hence,the dimensions of the pen that will maximize the area are L=160m and W=76800/160=480m

and we're done!