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Maximiser l'aire rectangulaire avec une clôture

Problème d'optimisation pour maximiser l'aire d'une clôture rectangulaire contre un mur de grange en utilisant le calcul et les fonctions quadratiques.

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Problem

A farmer builds a rectangular pen against a barn wall with 300300 feet of fencing, using three fenced sides with bottom length xx and two equal ends of length yy; what dimensions maximize the area?

Step 1: Write the fencing constraint

Since only three sides need fencing, the total fencing gives

x+2y=300.x + 2y = 300.

Solving for yy in terms of xx gives

y=300x2.y = \frac{300 - x}{2}.

Step 2: Express area as a function of one variable

The area of the rectangle is

A=xy.A = xy.

Substitute y=300x2y = \dfrac{300-x}{2} to get

A(x)=x300x2=150xx22.A(x) = x \cdot \frac{300-x}{2} = 150x - \frac{x^2}{2}.

This is a downward-opening parabola in xx.

Step 3: Differentiate and find the peak

Differentiate the area function:

A(x)=150x.A'(x) = 150 - x.

Set the derivative equal to 00:

150x=0.150 - x = 0.

So

x=150.x = 150.

Then

y=3001502=75.y = \frac{300 - 150}{2} = 75.

Step 4: Compute the maximum area

The maximizing dimensions are x=150x = 150 feet and y=75y = 75 feet, so the maximum area is

15075=11,250150 \cdot 75 = 11{,}250

square feet.

Answer

The maximum area is 11,25011{,}250 square feet, achieved when the pen measures 150150 feet by 7575 feet.

Concepts

Optimization Problems

Using derivatives to find the maximum or minimum value of a quantity in a real-world context. Set up an objective function from the problem, find its critical points, and verify whether each is a maximum or minimum.

Quadratic Functions and Graphs

Quadratic functions and their parabolic graphs. Can be written in standard form y=ax2+bx+cy = ax^2 + bx + c, vertex form y=a(xh)2+ky = a(x - h)^2 + k, or factored form y=a(xr1)(xr2)y = a(x - r_1)(x - r_2). The vertex, axis of symmetry, direction of opening, and intercepts describe the parabola.

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