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Sagot :
Part (A): Finding a Function for the Area [tex]\( A(x) \)[/tex]
We need to find a function for the area [tex]\( A(x) \)[/tex] of the garden in terms of [tex]\( x \)[/tex] , where [tex]\( x \)[/tex] is the length of the garden parallel to the house.
1. Define Variables:
- Let [tex]\( x \)[/tex] be the length of the garden parallel to the house.
- Let [tex]\( y \)[/tex] be the width of the garden perpendicular to the house.
2. Set Up the Equation for the Fencing:
The fencing is used for three sides of the garden (two widths and one length):
[tex]\[ x + 2y = 320 \][/tex]
3. Solve for [tex]\( y \):[/tex]
[tex]\[ 2y = 320 - x \] \[ y = \frac{320 - x}{2} \][/tex]
4. Express the Area in Terms of \( x \):
The area [tex]\( A \)[/tex] of the rectangle is:
[tex]\[ A = x \times y \][/tex]
Substitute [tex]\( y \)[/tex] from the previous step:
[tex]\[ A(x) = x \left(\frac{320 - x}{2}\right) \] \[ A(x) = \frac{x(320 - x)}{2} \] \[ A(x) = \frac{320x - x^2}{2} \][/tex]
Thus, the function for the area [tex]\( A(x) \)[/tex] is:
[tex]\[ A(x) = \frac{320x - x^2}{2} \][/tex]
Part (B): Finding the Side Length [tex]\( x \)[/tex] that Gives the Maximum Area
To find the maximum area, we use the properties of quadratic functions. The function[tex]\( A(x) = \frac{320x - x^2}{2} \)[/tex] is a quadratic function of the form [tex]\( A(x) = -\frac{1}{2}x^2 + 160x \).[/tex]
For a quadratic function[tex]\( ax^2 + bx + c \),[/tex] the maximum or minimum value occurs at the vertex, and the [tex]\( x \)[/tex] -coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
1. Identify Coefficients:
For the quadratic function [tex]\( A(x) = -\frac{1}{2}x^2 + 160x \):[/tex]
-[tex]\( a = -\frac{1}{2} \)[/tex]
- [tex]\( b = 160 \)[/tex]
2. Calculate the Vertex:
[tex]\[ x = -\frac{b}{2a} \] \[ x = -\frac{160}{2 \times -\frac{1}{2}} \] \[ x = -\frac{160}{-1} \] \[ x = 160 \][/tex]
Thus, the side length [tex]\( x \)[/tex] that gives the maximum area is:
[tex]\[ x = 160 \text{ meters} \][/tex]
Part (C): Finding the Maximum Area
Substitute[tex]\( x = 160 \)[/tex] back into the area function [tex]\( A(x) \)[/tex] to find the maximum area.
1. Substitute [tex]\( x = 160 \) into \( A(x) \)[/tex]:
[tex]\[ A(160) = \frac{320 \times 160 - 160^2}{2} \] \[ A(160) = \frac{51200 - 25600}{2} \] \[ A(160) = \frac{25600}{2} \] \[ A(160) = 12800 \text{ square meters} \][/tex]
(A) The function for the area [tex]\( A(x) \)[/tex] in terms of [tex]\( x \)[/tex] is:
[tex]\[ A(x) = \frac{320x - x^2}{2} \][/tex]
(B) The side length [tex]\( x \)[/tex] that gives the maximum area is:
[tex]\[ x = 160 \text{ meters} \][/tex]
(C) The maximum area that the garden can have is:
[tex]\[ 12800 \text{ square meters} \][/tex]
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