How do you find the dimensions of a rectangle whose area is 100 square meters and whose perimeter is a minimum?

Question

Cameron Alexander · Accepted Answer

Let #x# and #y# be the base and the height of the rectangle, respectively.

Since the area is 100 #m^2#,

#xy=100 Rightarrow y=100/x#

The perimeter #P# can be expressed as

#P=2(x+y)=2(x+100/x)#

So, we want to minimize #P(x)# on #(0,infty)#.

By taking the derivative,

#P'(x)=2(1-100/x^2)=0 Rightarrowx=pm10#

#x=10# is the only critical value on #(0,infty)#

#y=100/10=10#

By testing some sample values,

#P'(1)<0 Rightarrow P(x)# is decreasing on #(0,10]#.

#P'(11)>0 Rightarrow P(x)# is increasing on #[10,infty)#

Therefore, #P(10)# is the minimum

I hope that this was helpful.

Hence, the dimensions are #10\times10#.

Scarlett Allison · Answer

undefined To find the dimensions of a rectangle with a given area of 100 square meters and a minimum perimeter, you would need to apply the concept of optimization. Let $ l $ represent the length of the rectangle and $ w $ represent the width. The formulas for area and perimeter of a rectangle are:

Area = Length $ \times $ Width
Perimeter = 2$ \times $ (Length + Width)

To minimize the perimeter while keeping the area constant, you can use the following steps:

1. Express one of the dimensions in terms of the other using the area formula.
2. Substitute the expression for one dimension into the formula for the perimeter.
3. Find the derivative of the resulting expression with respect to the remaining variable.
4. Set the derivative equal to zero and solve for the variable.
5. Use the obtained value to find the dimensions.

Given the area $ A = 100 $ square meters:

1. Express one dimension in terms of the other: $ l = \frac{100}{w} $.
2. Substitute into the perimeter formula: $ P = 2(l + w) = 2\left(\frac{100}{w} + w\right) $.
3. Find the derivative of the perimeter with respect to width: $ \frac{dP}{dw} = -\frac{200}{w^2} + 2 $.
4. Set the derivative equal to zero and solve for width: $ -\frac{200}{w^2} + 2 = 0 $.
5. Solve for $ w $: $ w^3 = 100 \Rightarrow w = \sqrt[3]{100} $.
6. Find the corresponding length using the formula for length: $ l = \frac{100}{w} $.

Thus, the dimensions of the rectangle that minimize its perimeter are approximately $ w \approx 4.64 $ meters (width) and $ l \approx 21.54 $ meters (length).