Let's solve this problem step-by-step.
The original dimensions of the pumpkin patch are:
- Width = 40 meters
- Length = 60 meters
Joe's plans to expand each dimension are:
- Width increase = [tex]\(3x\)[/tex] meters
- Length increase = [tex]\(5x\)[/tex] meters
So, the new width and length of the pumpkin patch can be represented as:
- New width = [tex]\(40 + 3x\)[/tex] meters
- New length = [tex]\(60 + 5x\)[/tex] meters
We need to find the area of the new pumpkin patch as a function of [tex]\(x\)[/tex].
The area [tex]\(A(x)\)[/tex] of a rectangle is given by the product of its width and length:
[tex]\[A(x) = \text{New width} \times \text{New length}\][/tex]
[tex]\[A(x) = (40 + 3x) \times (60 + 5x)\][/tex]
To expand this product, we will use the distributive property of multiplication over addition:
[tex]\[
A(x) = (40 + 3x)(60 + 5x)
\][/tex]
Expanding this product:
[tex]\[
= 40 \cdot 60 + 40 \cdot 5x + 3x \cdot 60 + 3x \cdot 5x
\][/tex]
[tex]\[
= 2400 + 200x + 180x + 15x^2
\][/tex]
Combine like terms (the terms involving [tex]\(x\)[/tex]):
[tex]\[
A(x) = 2400 + 380x + 15x^2
\][/tex]
Therefore, the function [tex]\(A(x)\)[/tex] that represents the area of the new pumpkin patch in square meters is:
[tex]\[
A(x) = 15x^2 + 380x + 2400
\][/tex]
Now, let's match this result with the given options.
A. [tex]\(f(x)=15 x^2+380 x+2,400\)[/tex]
B. [tex]\(f(x)=15 x^2+420 x+2,400\)[/tex]
C. [tex]\(f(x)=15 x^2\)[/tex]
D. [tex]\(f(x)=15 x^2+2,400\)[/tex]
Clearly, option A is the correct match:
[tex]\[
f(x) = 15x^2 + 380x + 2400
\][/tex]
Thus, the correct answer is:
A. [tex]\(f(x)=15 x^2+380 x+2,400\)[/tex]
