Let length be L and width of open end be W. Height is then 144/(LW)
Area of netting required
A = LW (top) + WH (end) + 2LH (sides)
A = LW + 144/L + 288/W
Differentiate to find dA/dL and dA/dW and equate both to zero. Solve these as a pair of simultaneous equations. You should end up with L = 4.16m, W = 8.32m, H = 4.16m (2 dec. pl.)
a net enclosurefor practisinggolf shots is open at one end, as shown, find the dimensions that will minimize the amount of netting needed and give a volume of 144 m^3(netting is required only the sides, the top, the far end.)
2 answers
Find 2 formulas that will use the dimensions, i.e. surface area and volume.
Let x be width and height (since has a square face)
Let y be length of the rect. prism
so you have 3 rectangles (bottom surface ignored) and a square at the far end.
S.A. (surface area) = w*h+ 3(l*w)
= x^2 + 3xy
=x(x+3y)
Vol. = l*w*h
144m^3 =x*x*y
rearrange vol. for y
144/x^2 = y
plug y into your S.A. equation and solve for x and simplify
Now find where the critical point (C.P.) of your S.A. formula using the 1st derivative of your new S.A. equation
S.A.' = 2x - 432/x^2
C.P. @ S.A.' = 0
0 = 2x - 432/x^2
x = 6
take your vol. equ that you isolated y and solve for y with your new x
y = 144/x^2
y = 144/(6)^2
y = 4
Thus the dimensions are 4*6*6 (L*W*H)
Let x be width and height (since has a square face)
Let y be length of the rect. prism
so you have 3 rectangles (bottom surface ignored) and a square at the far end.
S.A. (surface area) = w*h+ 3(l*w)
= x^2 + 3xy
=x(x+3y)
Vol. = l*w*h
144m^3 =x*x*y
rearrange vol. for y
144/x^2 = y
plug y into your S.A. equation and solve for x and simplify
Now find where the critical point (C.P.) of your S.A. formula using the 1st derivative of your new S.A. equation
S.A.' = 2x - 432/x^2
C.P. @ S.A.' = 0
0 = 2x - 432/x^2
x = 6
take your vol. equ that you isolated y and solve for y with your new x
y = 144/x^2
y = 144/(6)^2
y = 4
Thus the dimensions are 4*6*6 (L*W*H)