Determine the zeroes of the quadratic function whose graphs are given below:
1 answer:
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Square means L=W
V=62.5=LWH
L=W so
V=62.5=HL^2
SA=2(L^2+2LH)
we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation
SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2
H=62.5/L^2
H=![\frac{25 \sqrt[3]{2} }{4}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%20%5Csqrt%5B3%5D%7B2%7D%20%7D%7B4%7D%20)
dimentions are
L=2.5∛4
W=2.5∛4
H=
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2
V=62.5=LWH
L=W so
V=62.5=HL^2
SA=2(L^2+2LH)
we have
V=62.5=HL^2
solve for H
divide both sides by L^2
62.5/L^2=H
sub that for H in other equation
SA=2(L^2+2L(62.5/L^2))
SA=2(L^2+125/L)
SA=2L^2+250/L
find minimum of 2L^2+250L^-1
take the derivitive
4L-250L^-2, or 2(2L^3-125)/L^2
find where it equals zero
it equals zero at L=2.5∛4
L=W
if we evaluate 2L^2+250/L at L=2.5∛4, the value is 75∛2
H=62.5/L^2
H=
dimentions are
L=2.5∛4
W=2.5∛4
H=
minimum surface area is 75∛2 in^2 or aprox 94.4941 in^2