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Answer :
To extend the storage lot by 115 feet in length while maintaining the same square footage, the additional width needed is half of the current width.
To determine the width of the additional space needed to extend the storage lot, we can use the information provided.
Let's denote:
- W as the current width of the storage lot.
- L as the current length of the storage lot.
- A as the additional length of the storage lot.
Given:
- The current storage lot can only be extended by 115 feet in length. So, A = 115 feet.
- We need to find the additional width (x) needed to maintain the same square footage.
Since the total area of the storage lot must remain the same after the extension, we can set up the equation:
[tex]\[ (W + x) \times (L + A) = W \times L \][/tex]
Substituting the given values:
[tex]\[ (W + x) \times (L + 115) = W \times L \][/tex]
[tex]\[ (W + x) \times L + (W + x) \times 115 = W \times L \][/tex]
[tex]\[ W \times L + x \times L + W \times 115 + x \times 115 = W \times L \][/tex]
[tex]\[ x \times L + W \times 115 + x \times 115 = 0 \][/tex]
Since [tex]W \times L[/tex] appears on both sides, it cancels out:
[tex]\[ x \times L + W \times 115 + x \times 115 = 0 \][/tex]
[tex]\[ x \times 115 + x \times L + W \times 115 = 0 \][/tex]
[tex]\[ x \times (115 + L) + W \times 115 = 0 \][/tex]
Now, we solve for \( x \):
[tex]\[ x \times (115 + L) = - W \times 115 \][/tex]
[tex]\[ x = \frac{- W \times 115}{115 + L} \][/tex]
Given that L = 115 and A = 115, substituting these values into the equation, we get:
[tex]\[ x = \frac{- W \times 115}{115 + 115} \][/tex]
[tex]\[ x = \frac{- W \times 115}{230} \][/tex]
[tex]\[ x = - \frac{W}{2} \][/tex]
Since the width cannot be negative, we take the positive value:
[tex]\[ x = \frac{W}{2} \][/tex]
Therefore, to extend the storage lot by 115 feet in length while maintaining the same square footage, the additional width needed is half of the current width.
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