Example 4: A right triangle whose hypotenuse is √48 inches long is revolved about one of its legs to generate a right circular cone (see figure). Find the lengths of the other two sides of the triangle that will maximize the volume of the cone and find the maximum volume that will occur. √√48

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Chapter1: Functions And Models
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Topic: Optimization Problems

Problem #4: 

A box with an open top is to be constructed from a 10 inch by 16 inch piece of cardboard by cutting squares of equal sides length from the corners and folding up the sides. Find the dimensions of the box of largest volume that can be constructed. 

4 problems in total that I need help with. 

**Mathematics Problem: Fencing a Rectangular Plot**

**Problem Description:**

A rectangular plot of land is to be fenced using two types of fencing. Two opposite sides will use heavy-duty fencing selling for $3 a foot. The two remaining sides will use standard fencing selling for $2 a foot. How much of the heavy-duty and standard fencing should be used so that the greatest area can be fenced in at a cost of $6,000?

**Visual Explanation:**

1. **Diagram of the Plot:**
   - A rectangle is drawn to represent the plot.
   - The horizontal sides are labeled as "x ft" and the vertical sides as "y ft."

2. **Formulas and Definitions:**
   - \( x = \text{price per foot for the heavy-duty fencing} \cdot 3 \, \text{dollars/foot} \)
   - \( y = \text{price per foot for the standard fencing} \cdot 2 \, \text{dollars/foot} \)
   - The equation for the area \( A \) is defined with respect to a total cost of $6,000.

3. **Solution Annotation:**
   - The maximum area is calculated to be 375,000 square feet.
   - The optimal use of fencing is indicated as 1,000 feet of standard fencing and 1,000 feet of heavy-duty fencing.
Transcribed Image Text:**Mathematics Problem: Fencing a Rectangular Plot** **Problem Description:** A rectangular plot of land is to be fenced using two types of fencing. Two opposite sides will use heavy-duty fencing selling for $3 a foot. The two remaining sides will use standard fencing selling for $2 a foot. How much of the heavy-duty and standard fencing should be used so that the greatest area can be fenced in at a cost of $6,000? **Visual Explanation:** 1. **Diagram of the Plot:** - A rectangle is drawn to represent the plot. - The horizontal sides are labeled as "x ft" and the vertical sides as "y ft." 2. **Formulas and Definitions:** - \( x = \text{price per foot for the heavy-duty fencing} \cdot 3 \, \text{dollars/foot} \) - \( y = \text{price per foot for the standard fencing} \cdot 2 \, \text{dollars/foot} \) - The equation for the area \( A \) is defined with respect to a total cost of $6,000. 3. **Solution Annotation:** - The maximum area is calculated to be 375,000 square feet. - The optimal use of fencing is indicated as 1,000 feet of standard fencing and 1,000 feet of heavy-duty fencing.
**Example 3:** Determine the side lengths to maximize the area of a rectangle with a perimeter of 100 meters.

**Example 4:** A right triangle whose hypotenuse is \(\sqrt{48}\) inches long is revolved about one of its legs to generate a right circular cone (see figure). Find the lengths of the other two sides of the triangle that will maximize the volume of the cone and find the maximum volume that will occur.

*Diagram Description:*

The diagram depicts a right circular cone formed by rotating a right triangle around one of its legs. The hypotenuse of the triangle has a length of \(\sqrt{48}\). The height and base of the cone (the two legs of the right triangle) are to be determined to maximize the cone's volume.
Transcribed Image Text:**Example 3:** Determine the side lengths to maximize the area of a rectangle with a perimeter of 100 meters. **Example 4:** A right triangle whose hypotenuse is \(\sqrt{48}\) inches long is revolved about one of its legs to generate a right circular cone (see figure). Find the lengths of the other two sides of the triangle that will maximize the volume of the cone and find the maximum volume that will occur. *Diagram Description:* The diagram depicts a right circular cone formed by rotating a right triangle around one of its legs. The hypotenuse of the triangle has a length of \(\sqrt{48}\). The height and base of the cone (the two legs of the right triangle) are to be determined to maximize the cone's volume.
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