This is a trivial question; but I just want to make sure:
A closed cylindrical container has a capacity of $128\pi \,{\rm m}^3$. Determine the minimum surface area. The answer is $96\pi$.
Volume of Cylinder, $V = \pi r^2 h = 128\pi$ (eq1)
Surface Area of Cylinder, $SA = 2\pi(r^2 + rh)$ (eq2)
Substitute eq(1) and eq(2); solve for the derivative of zero: $$\frac{\rm d}{{\rm d}r}( r^2 + 128/r) = 0,$$ solve for $r$.
We get $r=4$.
Put back into $V$ formula, $h = 2.54647$
Calculate Surface area: $52\pi$
Is it possible answer is wrong? I double checked with wolfram Alpha and all my derivatives are valid.
Video Transcript
The volume of the slender, the volume of the cylindrical tank, the volume of the sylintricaltank, is equal to 64 n cube final diameter of the cylindrical tank, which required the minimum amount of metal. So this is our. This is our slender. So this is our cylinders, which is open from top. So not the volume v is equal to pi r square h. So from here, pi r square h is equal to 64 point. The value of h is equal to 64 over pi r square. So this is the equation: number 1. Now the total surface area. The total surface area is equal to 2 pi r, h, plus the base area, that is, a pi r square soto. The value of h, here, 2 pi r multiplied to 64 over pi r square, plus pi r square pi pi get canceled out, so this is equal to 128 over r plus pi r square. So it implies the s of r is equal to this now differentiate. The it with with respect to r this equal to 2 pi r negative 128 over r square, now substitute a dash of r, is equal to 0. It implies 2 pi r, negative 128 over r square is equal to 0. Now, from here, the 2 pi r cubed is equal to 128 poi, so the pi r cube is equal to 64 point. So therefore, r c equal to 64 over pi, so from here value of r, is equal to 64 over pi to the whole power 1. Over 3 pi, so this is equal to so this equal to 2.731. Therefore, the radius of the cylindrical tank is equal to 2.731 inch now again differentiate the s with respect to r, so this is equal to 2 pi plus 256 over r cube. Now the value of double derivative at 2.731 is equal to 2 pi plus 256. Over 2.731 to the whole cube is greater than 0. So from here we can imply that the hat the surface area- hath minimum value at the value of radius r, is equal to 2.731. So, therefore, the diameter d is equal to twice of r, and this is equal to 2 multiplied to 2.731. So now this is equal to this is equal to 5.462 inch. It implies the diameter it implies. The diameter of the soletrical container is equal to 5.462 in so this is our this required finalanswer.
Video Transcript
The volume of the slender, the volume of the cylindrical tank, the volume of the sylintricaltank, is equal to 64 n cube final diameter of the cylindrical tank, which required the minimum amount of metal. So this is our. This is our slender. So this is our cylinders, which is open from top. So not the volume v is equal to pi r square h. So from here, pi r square h is equal to 64 point. The value of h is equal to 64 over pi r square. So this is the equation: number 1. Now the total surface area. The total surface area is equal to 2 pi r, h, plus the base area, that is, a pi r square soto. The value of h, here, 2 pi r multiplied to 64 over pi r square, plus pi r square pi pi get canceled out, so this is equal to 128 over r plus pi r square. So it implies the s of r is equal to this now differentiate. The it with with respect to r this equal to 2 pi r negative 128 over r square, now substitute a dash of r, is equal to 0. It implies 2 pi r, negative 128 over r square is equal to 0. Now, from here, the 2 pi r cubed is equal to 128 poi, so the pi r cube is equal to 64 point. So therefore, r c equal to 64 over pi, so from here value of r, is equal to 64 over pi to the whole power 1. Over 3 pi, so this is equal to so this equal to 2.731. Therefore, the radius of the cylindrical tank is equal to 2.731 inch now again differentiate the s with respect to r, so this is equal to 2 pi plus 256 over r cube. Now the value of double derivative at 2.731 is equal to 2 pi plus 256. Over 2.731 to the whole cube is greater than 0. So from here we can imply that the hat the surface area- hath minimum value at the value of radius r, is equal to 2.731. So, therefore, the diameter d is equal to twice of r, and this is equal to 2 multiplied to 2.731. So now this is equal to this is equal to 5.462 inch. It implies the diameter it implies. The diameter of the soletrical container is equal to 5.462 in so this is our this required finalanswer.
Question:
A cylindrical container must hold 2 L or 2,000 cm{eq}^3 {/eq} of liquid. Find the dimensions of the container which will minimize the amount of material needed.
a. r = 6.83 cm; h = 13.65 cm
b. r = 13.65 cm; h = 6.83 cm
c. r = 17.84 cm; h = 2.00 cm
d. r = 6.83 cm; h = 6.83 cm
Optimizing Functions using Differential Calculus
A function can be optimized by using its derivatives from differential calculus. By the definition of maxima and minima, the first derivative of the function at these points will be zero. The second derivative test can be used to determine if the extremum is a relative maximum or minimum.
Answer and Explanation:
Given: A cylindrical container that must hold a volume {eq}V = 2000\ cm^3 {/eq}.
Assume a closed cylinder with base of radius {eq}r {/eq} and...
See full answer below.
Learn more about this topic:
Finding Minima & Maxima: Problems & Explanation
from
Chapter 5 / Lesson 2
One of the most important practical uses of higher mathematics is finding minima and maxima. This lesson will describe different ways to determine the maxima and minima of a function and give some real world examples.