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You have land that you would like to use to create two distinct fenced-in areas in the shape given below. You have 410 meters of fencing materials to use. What values of x and y would result in the maximum area that you can enclose

Sagot :

MrRoyalidn

Answer:

[tex]x = \frac{205}{3}[/tex]

[tex]y =\frac{195}{4}[/tex]

Step-by-step explanation:

Given

[tex]p = 410[/tex] --- perimeter

See attachment for fence

Required

x and y

The perimeter of the fence is:

[tex]p = 2(x + y +5 + y) +x[/tex]

Open bracket

[tex]p = 2x + 2y +10 + 2y +x[/tex]

Collect like terms

[tex]p = 2x+x + 2y + 2y+10[/tex]

[tex]p = 3x + 4y+10[/tex]

Substitute: [tex]p = 410[/tex]

[tex]3x + 4y+10 =410[/tex]

Make 4y the subject

[tex]4y =410-10-3x[/tex]

[tex]4y =400-3x[/tex]

Make y the subject

[tex]y =\frac{400-3x}{4}[/tex]

The area (A) of the fence is:

[tex]A = (y + y + 5) * x[/tex]

[tex]A = (2y + 5) * x[/tex]

Substitute: [tex]y =\frac{400-3x}{4}[/tex]

[tex]A = (2*\frac{400-3x}{4} + 5) * x[/tex]

[tex]A = (\frac{400-3x}{2} + 5) * x[/tex]

Take LCM

[tex]A = (\frac{400-3x+10}{2}) * x[/tex]

Solve like terms

[tex]A = (\frac{410-3x}{2}) * x[/tex]

Open bracket

[tex]A = \frac{410x-3x^2}{2}[/tex]

Remove fraction

[tex]A = 205x-1.5x^2[/tex]

Differentiate both sides

[tex]A' = 205 - 3x[/tex]

To maximize; set [tex]A' =0[/tex]

[tex]205 - 3x =0[/tex]

Solve for 3x

[tex]3x = 205[/tex]

Solve for x

[tex]x = \frac{205}{3}[/tex]

Recall that: [tex]y =\frac{400-3x}{4}[/tex]

So, we have:

[tex]y =\frac{400-3*205/3}{4}[/tex]

[tex]y =\frac{400-205}{4}[/tex]

[tex]y =\frac{195}{4}[/tex]

View image MrRoyal
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