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A solar oven is to be made from an open box with reflective sides. Each box is made from a 30-in by 16-in rectangular sheet of aluminum with squares of length x (in inches) removed from each corner. Then the flaps are folded up to form an open box. 16 in 30 in 16 - 2x Part 1 of 5 (a) Show that the volume of the box is given by V(x) = 4x -92x+480x for 0

A solar oven is to be made from an open box with reflective sides. Each box is made from a 30-in by 16-in rectangular sheet of aluminum with squares of length x (in inches) removed from each corner. Then the flaps are folded up to form an open box. 16 in 30 in 16 - 2x Part 1 of 5 (a) Show that the volume of the box is given by V(x) = 4x -92x+480x for 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
A solar oven is to be made from an open box with reflective sides. Each box is made from a 30-in by 16-in rectangular sheet of aluminum with squares of length x (in inches) removed from each corner. Then the flaps are folded up to form an open box. 16 in 30 in 16 - 2r Part 1 of 5 (a) Show that the volume of the box is given by V (x) = 4x - 92x +480x for 0<x<8. The formula for the volume of a box is V= Iwh Part 2 of 5 If the height of the box is x, the expression for its length is 30 - 2x, and the expression for its width is 16 - 2x Part 3 of 5 Therefore, V (x) = (( 30 – 2x)( 16 – 2x)(x), which simplifies to V (x) = 4x - 92x+ 480x for 0 <x< 8. Part: 3 /5 Part 4 of 5 (b) Graph the function from part (a) and use the "Maximum" feature on a graphing utility to approximate the length of the sides of the squares that should be removed to maximize the volume. Round to the nearest tenth of an inch. The length of the sides of the squares that should be removed to maximize the volume is approximately in.
Transcribed Image Text:A solar oven is to be made from an open box with reflective sides. Each box is made from a 30-in by 16-in rectangular sheet of aluminum with squares of length x (in inches) removed from each corner. Then the flaps are folded up to form an open box. 16 in 30 in 16 - 2r Part 1 of 5 (a) Show that the volume of the box is given by V (x) = 4x - 92x +480x for 0<x<8. The formula for the volume of a box is V= Iwh Part 2 of 5 If the height of the box is x, the expression for its length is 30 - 2x, and the expression for its width is 16 - 2x Part 3 of 5 Therefore, V (x) = (( 30 – 2x)( 16 – 2x)(x), which simplifies to V (x) = 4x - 92x+ 480x for 0 <x< 8. Part: 3 /5 Part 4 of 5 (b) Graph the function from part (a) and use the "Maximum" feature on a graphing utility to approximate the length of the sides of the squares that should be removed to maximize the volume. Round to the nearest tenth of an inch. The length of the sides of the squares that should be removed to maximize the volume is approximately in.
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