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dybincka [34]
3 years ago
9

A rectangular dog pen is constructed using a barn wall as one side and 60 meters of fencing for the other three sides. What is t

he maximum area of the dog pen?
Mathematics
1 answer:
irinina [24]3 years ago
8 0

Answer:

450 m²

Step-by-step explanation:

Let x = length of rectangle and y = width of the rectangle.

Therefore,

2x + y = 60...............(1)

Making y subject of the formula, we have:

y = 60 - 2x............... (2)

We know area of a rectangle is length * width i.e A = x*y

Let's substitute (60-2x) for y,

A = x * (60-2x)

= 60x - 2(x)²

= -2x² + 60x

\frac{dA}{dx} = -2x^2 + 60x =

-4x + 60 = 0

-4x = -60

x = \frac{-60}{-4} = 15

Let's substitute 15 for x in equation 2, we have:

y = 60 - 2(15)

= 60 - 30

y = 30

Since x = 15 & y = 30, the area would be:

A = xy

A = 15 * 30

A = 450 m²

The maximum area of the dog pen is 450m²

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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

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3 years ago
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<u>ANSWER</u>

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Therefore the parabola shown facing up with a vertex (2.5, -6) is not a one-to-one function.

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