If A Company Is Producing A Rectangular Pen Holder With A Fixed Perimeter Of 20 Cm, And Wants To Maximize The Volume Of The Holder, But Also Has A Constraint That The Length Of The Holder Must Be At Least 2 Cm Longer Than The Width, How Can I Use Optimization Techniques To Find The Dimensions That Will Yield The Maximum Volume, Given That The Volume V Of The Holder Is Given By V = X * Y * Z, Where X, Y, And Z Are The Length, Width, And Height Of The Holder, Respectively, And The Height Is Fixed At 5 Cm?

To find the dimensions that maximize the volume of a rectangular pen holder with a fixed perimeter of 20 cm and a constraint that the length must be at least 2 cm longer than the width, we start by noting that the volume ${ V }$ is given by ${ V = x \cdot y \cdot z }$, where ${ x }$, ${ y }$, and ${ z }$ are the length, width, and height, respectively. The height ${ z }$ is fixed at 5 cm, so we need to maximize the area of the base ${ A = x \cdot y }$.

Given the fixed perimeter, we have: ${ 2x + 2y = 20 }$ which simplifies to: ${ x + y = 10 }$ Thus, we can express ${ y }$ in terms of ${ x }$: ${ y = 10 - x }$

We also have the constraint that the length must be at least 2 cm longer than the width: ${ x \geq y + 2 }$ Substituting ${ y = 10 - x }$ into this inequality, we get: ${ x \geq (10 - x) + 2 }$ ${ x \geq 12 - x }$ ${ 2x \geq 12 }$ ${ x \geq 6 }$

Therefore, ${ x }$ must be at least 6 cm. Substituting ${ x = 6 }$ into ${ y = 10 - x }$, we get: ${ y = 10 - 6 = 4 }$

The area ${ A }$ is then: ${ A = x \cdot y = 6 \cdot 4 = 24 \, \text{cm}^2 }$

Since the height ${ z }$ is fixed at 5 cm, the volume ${ V }$ is: ${ V = 5 \cdot 24 = 120 \, \text{cm}^3 }$

Thus, the dimensions that yield the maximum volume are:

• Length: ${ \boxed{6} }$ cm
• Width: ${ \boxed{4} }$ cm
• Height: ${ \boxed{5} }$ cm