Seascapes Inc. wishes to build an open-top rectangular aquarium with a volume of

Question

Seascapes Inc. wishes to build an open-top rectangular aquarium with a volume of 12 cubic feet. The material for the bottom costs $6 per square foot, and the material for the sides cost $2 per square foot. (Remember no top). Find the dimensions of the least expensive aquarium Seascapes Inc. can build. A) What is to be optimized? _________ Now use the variables x, y. z to write an equation for the item to be optimized. __________ B) Changes in what cause changes in A? _____________ C) Find (A) in terms of (B): i) Is there a catch? What is it? ____________ ii) Use the catch to write the equation from (A) in terms of two variables only. ____________ D) Find the critical point of Cii and show all the correct work needed to do so. _________________ E) Test the critical point you found in part (D) to verify it is a minimum. (show work!) ______________ F) Think!! Do your answers to parts (D) and (E) make sense? G) State the solution using a complete sentence. _________________

Explanation / Answer

(A) HEre the cost of the material have to be optimized.

If the dimension of the aquarium are x,y and Z foot. (LEngth , width and breadth) respectivly.

Then cost of bottom = 6xy (as there is no top we will not include it)

cost of 4 sides = 2 * 2(xz + yz) = 4z (x +y)

so total cost A = 6xy + 4z ( x+y)

(b) changes in dimensions of cube will put changes in (A)

(C) (i) Here Volume of cube is given, wich is equal to 12 ft cube

that means V = xyz = 12 ft cube

(ii) so here we can write the equation in 2 variable only as z = 12/xy

so A = 6xy + 4 * (12/xy) * (x+y)

A= 6xy + 48 (x+y)/xy

(D) Differentiate A w.r.t to x

dA/dx = 6Y + 48 * (-1/X2 ) ..(i)

dA/dy = 6X + 48 * (-1/Y2 ) /...(ii)

for optimum results dA/dx = dA/dy = 0

Y - 8/X2 = X - 8/Y2 = 0

X= (8)1/3 = 2 ft

Y = 321/3 = 2 ft

Z= 12/XY = 12/ 4 = 3 ft

(E) The critical point we get is (2,2,#)

so if we double differrentiate A with respect to x and y

we get d2 A/dx2 = - 96x-3 at x = 2, it has a negative value so double differentiation has a negative value that means it is a maxima here.

(F) Yes, our answrs ar D and E makes sense.

(G) We can conclude here that for a definite volumer of 12 ft3, we can have 2 ft of length , 2ft of width and 3 ft of heaight of the aquarium to minimizse its minimum cost of mateeiral.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.