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A contractor is to build a new structure at a farm for seed storage. It will have a rectangular base, which is 20 feet wide by 35 feet long, with walls of thickness t feet. The structure is 90 feet tall. Excluding the volume of the walls, which of the following functions best models the volume, V, in cubic feet, inside the structure? Choose 1 answer: (A) V(t)=63,000t^(2) (B) V(t)=35t^(2)+20 t+90 (C) V(t)=90(35-2t)(20-2t) (D) V(t)=90(35-2t)^(2)+20

A contractor is to build a new structure at a farm for seed storage. It will have a rectangular base, which is 2020 feet wide by 3535 feet long, with walls of thickness t t feet. The structure is 9090 feet tall. Excluding the volume of the walls, which of the following functions best models the volume, V V , in cubic feet, inside the structure?\newlineChoose 11 answer:\newline(A) V(t)=63,000t2 V(t)=63,000 t^{2} \newline(B) V(t)=35t2+20t+90 V(t)=35 t^{2}+20 t+90 \newline(C) V(t)=90(35−2t)(20−2t) V(t)=90(35-2 t)(20-2 t) \newline(D) V(t)=90(35−2t)2+20 V(t)=90(35-2 t)^{2}+20

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Q. A contractor is to build a new structure at a farm for seed storage. It will have a rectangular base, which is 2020 feet wide by 3535 feet long, with walls of thickness t t feet. The structure is 9090 feet tall. Excluding the volume of the walls, which of the following functions best models the volume, V V , in cubic feet, inside the structure?\newlineChoose 11 answer:\newline(A) V(t)=63,000t2 V(t)=63,000 t^{2} \newline(B) V(t)=35t2+20t+90 V(t)=35 t^{2}+20 t+90 \newline(C) V(t)=90(35−2t)(20−2t) V(t)=90(35-2 t)(20-2 t) \newline(D) V(t)=90(35−2t)2+20 V(t)=90(35-2 t)^{2}+20
  1. Calculate Volume: First, calculate the volume of the structure without considering the walls. The formula for the volume of a rectangular prism is V=length×width×heightV = \text{length} \times \text{width} \times \text{height}.V=35 feet×20 feet×90 feetV = 35 \, \text{feet} \times 20 \, \text{feet} \times 90 \, \text{feet}
  2. Perform Multiplication: Now, perform the multiplication to find the volume. V=63,000V = 63,000 cubic feet
  3. Account for Walls: Next, we need to account for the thickness of the walls, which will reduce the interior volume. The walls take up space on all sides, so we subtract twice the thickness of the walls from both the length and the width.\newlineThe new length is (35−2t)(35 - 2t) feet, and the new width is (20−2t)(20 - 2t) feet.
  4. Write Volume Function: Now, write the function for the volume inside the structure, excluding the walls.\newlineV(t)=height×(length−2×thickness)×(width−2×thickness)V(t) = \text{height} \times (\text{length} - 2 \times \text{thickness}) \times (\text{width} - 2 \times \text{thickness})\newlineV(t)=90×(35−2t)×(20−2t)V(t) = 90 \times (35 - 2t) \times (20 - 2t)
  5. Check Answer Choices: Check the answer choices to see which one matches the derived function.\newlineV(t)=90(35−2t)(20−2t)V(t) = 90(35 - 2t)(20 - 2t) matches with option (C).

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