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### Question 13:
A farmer has 1200 meters of wire to fence three sides of a rectangular field, with the fourth side being a wall. We need to determine the dimensions of the field that will give the maximum possible area.
Step-by-Step Solution:
1. Define variables:
- Let [tex]\( \text{length} \)[/tex] be the length of the field parallel to the wall.
- Let [tex]\( \text{width} \)[/tex] be the width of the field perpendicular to the wall.
2. Write the equation for the perimeter:
- Since the wire is used to fence three sides, we have one length and two widths:
[tex]\[ \text{length} + 2 \times \text{width} = 1200 \][/tex]
3. Isolate the length:
- We can solve for [tex]\(\text{length}\)[/tex] in terms of [tex]\(\text{width}\)[/tex]:
[tex]\[ \text{length} = 1200 - 2 \times \text{width} \][/tex]
4. Express the area:
- The area [tex]\(A\)[/tex] of the field is given by:
[tex]\[ A = \text{length} \times \text{width} = (1200 - 2 \times \text{width}) \times \text{width} - Substitute the expression for length: \[ A = 1200 \times \text{width} - 2 \times \text{width}^2 \][/tex]
5. Find the critical points:
- To maximize the area, we find the derivative of the area function with respect to width, set it to zero, and solve for [tex]\(\text{width}\)[/tex]:
[tex]\[ \frac{dA}{d\text{width}} = 1200 - 4 \times \text{width} = 0 \][/tex]
[tex]\[ 1200 = 4 \times \text{width} \][/tex]
[tex]\[ \text{width} = 300 \][/tex]
6. Determine the length:
- Substitute [tex]\(\text{width} = 300\)[/tex] back into the equation for the length:
[tex]\[ \text{length} = 1200 - 2 \times 300 = 600 \][/tex]
7. Calculate the maximum area:
- The maximum area is then:
[tex]\[ A = \text{length} \times \text{width} = 600 \times 300 = 180,000 \, \text{square meters} \][/tex]
Hence, the dimensions that give the maximum possible area are:
- Length: 600 meters
- Width: 300 meters
- Maximum Area: 180,000 square meters
### Question 14:
Given that quantity [tex]\(P\)[/tex] is partly constant and partly varies as the cube of [tex]\(Q\)[/tex], we need to find the value of [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex].
Step-by-Step Solution:
1. Express the relationship:
- Let [tex]\(P\)[/tex] have the form:
[tex]\[ P = k + mQ^3 \][/tex]
- Where [tex]\(k\)[/tex] is the constant part and [tex]\(mQ^3\)[/tex] is the part that varies as the cube of [tex]\(Q\)[/tex].
2. Use given conditions to form equations:
- When [tex]\(Q = 1\)[/tex], [tex]\(P = 23\)[/tex]:
[tex]\[ 23 = k + m(1^3) \implies 23 = k + m \][/tex]
- When [tex]\(Q = 2\)[/tex], [tex]\(P = 44\)[/tex]:
[tex]\[ 44 = k + m(2^3) \implies 44 = k + 8m \][/tex]
3. Solve the system of equations:
- We have two equations:
[tex]\[ \begin{cases} 23 = k + m \\ 44 = k + 8m \end{cases} \][/tex]
- Subtract the first equation from the second equation:
[tex]\[ (44 - 23) = (k + 8m) - (k + m) \implies 21 = 7m \implies m = 3 \][/tex]
- Substitute [tex]\(m = 3\)[/tex] back into the first equation:
[tex]\[ 23 = k + 3 \implies k = 20 \][/tex]
4. Determine [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex]:
- Use the formula with the found values of [tex]\(k\)[/tex] and [tex]\(m\)[/tex]:
[tex]\[ P = 20 + 3(5^3) = 20 + 3(125) = 20 + 375 = 395 \][/tex]
Thus, the value of [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex] is:
[tex]\[ \boxed{395} \][/tex]
A farmer has 1200 meters of wire to fence three sides of a rectangular field, with the fourth side being a wall. We need to determine the dimensions of the field that will give the maximum possible area.
Step-by-Step Solution:
1. Define variables:
- Let [tex]\( \text{length} \)[/tex] be the length of the field parallel to the wall.
- Let [tex]\( \text{width} \)[/tex] be the width of the field perpendicular to the wall.
2. Write the equation for the perimeter:
- Since the wire is used to fence three sides, we have one length and two widths:
[tex]\[ \text{length} + 2 \times \text{width} = 1200 \][/tex]
3. Isolate the length:
- We can solve for [tex]\(\text{length}\)[/tex] in terms of [tex]\(\text{width}\)[/tex]:
[tex]\[ \text{length} = 1200 - 2 \times \text{width} \][/tex]
4. Express the area:
- The area [tex]\(A\)[/tex] of the field is given by:
[tex]\[ A = \text{length} \times \text{width} = (1200 - 2 \times \text{width}) \times \text{width} - Substitute the expression for length: \[ A = 1200 \times \text{width} - 2 \times \text{width}^2 \][/tex]
5. Find the critical points:
- To maximize the area, we find the derivative of the area function with respect to width, set it to zero, and solve for [tex]\(\text{width}\)[/tex]:
[tex]\[ \frac{dA}{d\text{width}} = 1200 - 4 \times \text{width} = 0 \][/tex]
[tex]\[ 1200 = 4 \times \text{width} \][/tex]
[tex]\[ \text{width} = 300 \][/tex]
6. Determine the length:
- Substitute [tex]\(\text{width} = 300\)[/tex] back into the equation for the length:
[tex]\[ \text{length} = 1200 - 2 \times 300 = 600 \][/tex]
7. Calculate the maximum area:
- The maximum area is then:
[tex]\[ A = \text{length} \times \text{width} = 600 \times 300 = 180,000 \, \text{square meters} \][/tex]
Hence, the dimensions that give the maximum possible area are:
- Length: 600 meters
- Width: 300 meters
- Maximum Area: 180,000 square meters
### Question 14:
Given that quantity [tex]\(P\)[/tex] is partly constant and partly varies as the cube of [tex]\(Q\)[/tex], we need to find the value of [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex].
Step-by-Step Solution:
1. Express the relationship:
- Let [tex]\(P\)[/tex] have the form:
[tex]\[ P = k + mQ^3 \][/tex]
- Where [tex]\(k\)[/tex] is the constant part and [tex]\(mQ^3\)[/tex] is the part that varies as the cube of [tex]\(Q\)[/tex].
2. Use given conditions to form equations:
- When [tex]\(Q = 1\)[/tex], [tex]\(P = 23\)[/tex]:
[tex]\[ 23 = k + m(1^3) \implies 23 = k + m \][/tex]
- When [tex]\(Q = 2\)[/tex], [tex]\(P = 44\)[/tex]:
[tex]\[ 44 = k + m(2^3) \implies 44 = k + 8m \][/tex]
3. Solve the system of equations:
- We have two equations:
[tex]\[ \begin{cases} 23 = k + m \\ 44 = k + 8m \end{cases} \][/tex]
- Subtract the first equation from the second equation:
[tex]\[ (44 - 23) = (k + 8m) - (k + m) \implies 21 = 7m \implies m = 3 \][/tex]
- Substitute [tex]\(m = 3\)[/tex] back into the first equation:
[tex]\[ 23 = k + 3 \implies k = 20 \][/tex]
4. Determine [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex]:
- Use the formula with the found values of [tex]\(k\)[/tex] and [tex]\(m\)[/tex]:
[tex]\[ P = 20 + 3(5^3) = 20 + 3(125) = 20 + 375 = 395 \][/tex]
Thus, the value of [tex]\(P\)[/tex] when [tex]\(Q = 5\)[/tex] is:
[tex]\[ \boxed{395} \][/tex]
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